Physics Book - User contributions [en]

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2016-11-28T02:10:37Z

Zwright8: /* Path independence and round trip potential */

__NOTOC__
= '''Georgia Tech Student Wiki for Introductory Physics.''' =

This resource was created so that students can contribute and curate content to help those with limited or no access to a textbook. When reading this website, please correct any errors you may come across. If you read something that isn't clear, please consider revising it for future students!

Looking to make a contribution?
#Pick one of the topics from intro physics listed below
#Add content to that topic or improve the quality of what is already there.
#Need to make a new topic? Edit this page and add it to the list under the appropriate category. Then copy and paste the default [[Template]] into your new page and start editing.

Please remember that this is not a textbook and you are not limited to expressing your ideas with only text and equations. Whenever possible embed: pictures, videos, diagrams, simulations, computational models (e.g. Glowscript), and whatever content you think makes learning physics easier for other students.

== Source Material ==
All of the content added to this resource must be in the public domain or similar free resource. If you are unsure about a source, contact the original author for permission. That said, there is a surprisingly large amount of introductory physics content scattered across the web. Here is an incomplete list of intro physics resources (please update as needed).
* A physics resource written by experts for an expert audience [https://en.wikipedia.org/wiki/Portal:Physics Physics Portal]
* A wiki written for students by a physics expert [http://p3server.pa.msu.edu/coursewiki/doku.php?id=183_notes MSU Physics Wiki]
* A wiki book on modern physics [https://en.wikibooks.org/wiki/Modern_Physics Modern Physics Wiki]
* The MIT open courseware for intro physics [http://ocw.mit.edu/resources/res-8-002-a-wikitextbook-for-introductory-mechanics-fall-2009/index.htm MITOCW Wiki]
* An online concept map of intro physics [http://hyperphysics.phy-astr.gsu.edu/hbase/hph.html HyperPhysics]
* Interactive physics simulations [https://phet.colorado.edu/en/simulations/category/physics PhET]
* OpenStax algebra based intro physics textbook [https://openstaxcollege.org/textbooks/college-physics College Physics]
* The Open Source Physics project is a collection of online physics resources [http://www.opensourcephysics.org/ OSP]
* A resource guide compiled by the [http://www.aapt.org/ AAPT] for educators [http://www.compadre.org/ ComPADRE]

== Resources ==
* Commonly used wiki commands [https://en.wikipedia.org/wiki/Help:Cheatsheet Wiki Cheatsheet]
* A guide to representing equations in math mode [https://en.wikipedia.org/wiki/Help:Displaying_a_formula Wiki Math Mode]
* A page to keep track of all the physics [[Constants]]
* A page for review of [[Vectors]] and vector operations
* A listing of [[Notable Scientist]] with links to their individual pages

<div style="float:left; width:30%; padding:1%;">
==Physics 1==
===Week 1===
<div class="toccolours mw-collapsible mw-collapsed">
====Help with VPython====
<div class="mw-collapsible-content">
*[[VPython]]
*[[VPython basics]]
*[[VPython Common Errors and Troubleshooting]]
*[[VPython Functions]]
*[[VPython Lists]]
*[[VPython Loops]]
*[[VPython Multithreading]]
*[[VPython Animation]]
*[[VPython Objects]]
*[[VPython 3D Objects]]
*[[VPython Reference]]
*[[VPython MapReduceFilter]]
*[[VPython GUIs]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">

====Vectors and Units====
<div class="mw-collapsible-content">
*[[Vectors]]
*[[SI Units]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">

====Interactions====
<div class="mw-collapsible-content">
*[[Types of Interactions and How to Detect Them]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Velocity and Momentum====
<div class="mw-collapsible-content">
*[[Newton's First Law of Motion]]
*[[Velocity]]
*[[Mass]]
*[[Speed and Velocity]]
*[[Relative Velocity]]
*[[Derivation of Average Velocity]]
*[[2-Dimensional Motion]]
*[[3-Dimensional Position and Motion]]
</div>
</div>

===Week 2===
<div class="toccolours mw-collapsible mw-collapsed">
====Momentum and the Momentum Principle====
<div class="mw-collapsible-content">
*[[Momentum Principle]]
*[[Inertia]]
*[[Net Force]]
*[[Derivation of the Momentum Principle]]
*[[Impulse Momentum]]
*[[Acceleration]]
*[[Momentum with respect to external Forces]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Iterative Prediction with a Constant Force====
<div class="mw-collapsible-content">
*[[Newton’s Second Law of Motion]]
*[[Iterative Prediction]]
*[[Kinematics]]
*[[Newton’s Laws and Linear Momentum]]
*[[Projectile Motion]]
</div>
</div>

===Week 3===
<div class="toccolours mw-collapsible mw-collapsed">
====Analytic Prediction with a Constant Force====
<div class="mw-collapsible-content">
*[[Analytical Prediction]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Iterative Prediction with a Varying Force====
<div class="mw-collapsible-content">
*[[Predicting Change in multiple dimensions]]
*[[Spring Force]]
*[[Hooke's Law]]
*[[Simple Harmonic Motion]]
*[[Iterative Prediction of Spring-Mass System]]
*[[Terminal Speed]]
*[[Determinism]]
</div>
</div>

===Week 4===
<div class="toccolours mw-collapsible mw-collapsed">
====Fundamental Interactions====
<div class="mw-collapsible-content">
*[[Gravitational Force]]
*[[An Application of Gravitational Potential]]
*[[Electric Force]]
*[[Reciprocity]]
</div>
</div>

===Week 5===
<div class="toccolours mw-collapsible mw-collapsed">
====Conservation of Momentum====
<div class="mw-collapsible-content">
*[[Conservation of Momentum]]
</div>
</div>
<div class="toccolours mw-collapsible mw-collapsed">

====Properties of Matter====
<div class="mw-collapsible-content">
*[[Kinds of Matter]]
*[[Ball and Spring Model of Matter]]
*[[Density]]
*[[Length and Stiffness of an Interatomic Bond]]
*[[Young's Modulus]]
*[[Speed of Sound in Solids]]
*[[Malleability]]
*[[Ductility]]
*[[Weight]]
*[[Hardness]]
*[[Boiling Point]]
*[[Melting Point]]
</div>
</div>

===Week 6===
<div class="toccolours mw-collapsible mw-collapsed">
====Identifying Forces====
<div class="mw-collapsible-content">
*[[Free Body Diagram]]
*[[Inclined Plane]]
*[[Compression or Normal Force]]
*[[Tension]]
</div>
</div>
<div class="toccolours mw-collapsible mw-collapsed">

====Curving Motion====
<div class="mw-collapsible-content">
*[[Curving Motion]]
*[[Centripetal Force and Curving Motion]]
*[[Perpetual Freefall (Orbit)]]
</div>
</div>

===Week 7===
<div class="toccolours mw-collapsible mw-collapsed">
====Energy Principle====
<div class="mw-collapsible-content">
*[[The Energy Principle]]
*[[Conservation of Energy]]
*[[Kinetic Energy]]
*[[Work]]
*[[Power (Mechanical)]]
</div>
</div>

===Week 8===
<div class="toccolours mw-collapsible mw-collapsed">
====Work by Non-Constant Forces====
<div class="mw-collapsible-content">
*[[Work Done By A Nonconstant Force]]
</div>
</div>
<div class="toccolours mw-collapsible mw-collapsed">
====Potential Energy====
<div class="mw-collapsible-content">
*[[Potential Energy]]
*[[Potential Energy of Macroscopic Springs]]
*[[Spring Potential Energy]]
*[[Ball and Spring Model]]
*[[Gravitational Potential Energy]]
*[[Energy Graphs]]
*[[Escape Velocity]]
</div>
</div>

===Week 9===
<div class="toccolours mw-collapsible mw-collapsed">
====Multiparticle Systems====
<div class="mw-collapsible-content">
*[[Center of Mass]]
*[[Multi-particle analysis of Momentum]]
*[[Momentum with respect to external Forces]]
*[[Potential Energy of a Multiparticle System]]
*[[Work and Energy for an Extended System]]
*[[Internal Energy]]
**[[Potential Energy of a Pair of Neutral Atoms]]
</div>
</div>

===Week 10===
<div class="toccolours mw-collapsible mw-collapsed">
====Choice of System====
<div class="mw-collapsible-content">
*[[System & Surroundings]]
</div>
</div>
<div class="toccolours mw-collapsible mw-collapsed">
====Thermal Energy, Dissipation and Transfer of Energy====
<div class="mw-collapsible-content">
*[[Thermal Energy]]
*[[Specific Heat]]
*[[Heat Capacity]]
*[[Specific Heat Capacity]]
*[[First Law of Thermodynamics]]
*[[Second Law of Thermodynamics and Entropy]]
*[[Temperature]]
*[[Predicting Change]]
*[[Energy Transfer due to a Temperature Difference]]
*[[Transformation of Energy]]
*[[The Maxwell-Boltzmann Distribution]]
*[[Air Resistance]]
</div>
</div>
<div class="toccolours mw-collapsible mw-collapsed">
====Rotational and Vibrational Energy====
<div class="mw-collapsible-content">
*[[Translational, Rotational and Vibrational Energy]]
</div>
</div>

===Week 11===
<div class="toccolours mw-collapsible mw-collapsed">
====Different Models of a System====
<div class="mw-collapsible-content">
*[[Point Particle Systems]]
*[[Real Systems]]
</div>
</div>
<div class="toccolours mw-collapsible mw-collapsed">
====Models of Friction====
<div class="mw-collapsible-content">
*[[Friction]]
*[[Static Friction]]
</div>
</div>

===Week 12===
<div class="toccolours mw-collapsible mw-collapsed">
====Collisions====
<div class="mw-collapsible-content">
*[[Newton's Third Law of Motion]]
*[[Collisions]]
*[[Elastic Collisions]]
*[[Inelastic Collisions]]
*[[Maximally Inelastic Collision]]
*[[Head-on Collision of Equal Masses]]
*[[Head-on Collision of Unequal Masses]]
*[[Scattering: Collisions in 2D and 3D]]
*[[Rutherford Experiment and Atomic Collisions]]
*[[Coefficient of Restitution]]
</div>
</div>

===Week 13===
<div class="toccolours mw-collapsible mw-collapsed">
====Rotations====
<div class="mw-collapsible-content">
*[[Rotation]]
*[[Angular Velocity]]
*[[Eulerian Angles]]
</div>
</div>
<div class="toccolours mw-collapsible mw-collapsed">
====Angular Momentum====
<div class="mw-collapsible-content">
*[[Total Angular Momentum]]
*[[Translational Angular Momentum]]
*[[Rotational Angular Momentum]]
*[[The Angular Momentum Principle]]
*[[Angular Momentum Compared to Linear Momentum]]
*[[Angular Impulse]]
*[[Predicting the Position of a Rotating System]]
*[[Angular Momentum of Multiparticle Systems]]
*[[The Moments of Inertia]]
*[[Moment of Inertia for a cylinder]]
*[[Right Hand Rule]]
</div>
</div>

===Week 14===
<div class="toccolours mw-collapsible mw-collapsed">
====Analyzing Motion with and without Torque====
<div class="mw-collapsible-content">
*[[Torque]]
*[[Torque 2]]
*[[Systems with Zero Torque]]
*[[Systems with Nonzero Torque]]
*[[Torque vs Work]]
*[[Gyroscopes]]
</div>
</div>

===Week 15===
<div class="toccolours mw-collapsible mw-collapsed">
====Introduction to Quantum Concepts====
<div class="mw-collapsible-content">
*[[Bohr Model]]
*[[Energy graphs and the Bohr model]]
*[[Quantized energy levels]]
</div>
</div>
</div>

<div style="float:left; width:30%; padding:1%;">
==Physics 2==
===Week 1===
<div class="toccolours mw-collapsible mw-collapsed">
====3D Vectors====
<div class="mw-collapsible-content">
*[[Vectors]]
*[[Right-Hand Rule]]
*[[Right Hand Rule]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Electric field====
<div class="mw-collapsible-content">
*[[Electric Field]]
*[[Electric Field and Electric Potential]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">

====Electric force====
<div class="mw-collapsible-content">
*[[Electric Force]]
*[[Lorentz Force]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
Aniruddha Nadkarni

====Electric field of a point particle====
<div class="mw-collapsible-content">
*[[Point Charge]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">

====Superposition====
<div class="mw-collapsible-content">
*[[Superposition Principle]]
*[[Superposition principle]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Dipoles====
<div class="mw-collapsible-content">
*[[Electric Dipole]]
*[[Magnetic Dipole]]
</div>
</div>

===Week 2===
<div class="toccolours mw-collapsible mw-collapsed">
====Interactions of charged objects====
<div class="mw-collapsible-content">
*[[Electric Field]]
*[[Electric Potential]]
*[[Electric Force]]
*[[Lorentz Force]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Tape experiments====
<div class="mw-collapsible-content">
*[[Polarization]]
*[[Electric Polarization]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Polarization====
<div class="mw-collapsible-content">
*[[Polarization]]
*[[Electric Polarization]]
*[[Polarization of an Atom]]
</div>
</div>

===Week 3===
<div class="toccolours mw-collapsible mw-collapsed">
====Insulators====
<div class="mw-collapsible-content">
*[[Insulators]]
*[[Potential Difference in an Insulator]]
*[[Charged Conductor and Charged Insulator]]
*[[Charged conductor and charged insulator]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Conductors====
<div class="mw-collapsible-content">
*[[Conductivity]]
*[[Charge Transfer]]
*[[Resistivity]]
*[[Polarization of a conductor]]
*[[Charged Conductor and Charged Insulator]]
*[[Charged conductor and charged insulator]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Charging and discharging====
<div class="mw-collapsible-content">
*[[Charge Transfer]]
*[[Electrostatic Discharge]]
*[[Charged Conductor and Charged Insulator]]
*[[Charged conductor and charged insulator]]
</div>
</div>

===Week 4===
<div class="toccolours mw-collapsible mw-collapsed">
====Field of a charged rod====
<div class="mw-collapsible-content">
*[[Charged Rod]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Field of a charged ring/disk/capacitor====
<div class="mw-collapsible-content">
*[[Charged Ring]]
*[[Charged Disk]]
*[[Charged Capacitor]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Field of a charged sphere====
<div class="mw-collapsible-content">
*[[Charged Spherical Shell]]
*[[Field of a Charged Ball]]
</div>
</div>

===Week 5===
<div class="toccolours mw-collapsible mw-collapsed">
====Potential energy====
<div class="mw-collapsible-content">
*[[Potential Energy - Claimed by Janki Patel]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Electric potential====
<div class="mw-collapsible-content">
*[[Electric Potential]]
*[[Path Independence of Electric Potential]]
*[[Potential Difference Path Independence, claimed by Aditya Mohile]]
*[[Potential Difference in a Uniform Field]]
*[[Potential Difference of Point Charge in a Non-Uniform Field]]
</div>
</div>
<div class="toccolours mw-collapsible mw-collapsed">
====Sign of a potential difference====
<div class="mw-collapsible-content">
*[[Sign of a Potential Difference]]
</div>
</div>
<div class="toccolours mw-collapsible mw-collapsed">
====Potential at a single location====
<div class="mw-collapsible-content">
*[[Electric Potential]]
*[[Potential Difference at One Location]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">

====Path independence and round trip potential====
<div class="mw-collapsible-content">
*[[Path Independence of Electric Potential]]
*[[Potential Difference Path Independence, claimed by Aditya Mohile]]
</div>
</div>

===Week 6===
<div class="toccolours mw-collapsible mw-collapsed">
====Electric field and potential in an insulator====
<div class="mw-collapsible-content">
*[[Potential Difference in an Insulator]]
*[[Electric Field in an Insulator]]
</div>
</div>
<div class="toccolours mw-collapsible mw-collapsed">
====Moving charges in a magnetic field====
<div class="mw-collapsible-content">
*[[Magnetic Field]]
*[[Magnetic Force]]
*[[Lorentz Force]]
</div>
</div>
<div class="toccolours mw-collapsible mw-collapsed">
====Biot-Savart Law====
<div class="mw-collapsible-content">
*[[Biot-Savart Law]]
*[[Biot-Savart Law for Currents]]
</div>
</div>
<div class="toccolours mw-collapsible mw-collapsed">
====Moving charges, electron current, and conventional current====
<div class="mw-collapsible-content">
*[[Moving Point Charge]]
*[[Current]]
</div>
</div>

===Week 7===
<div class="toccolours mw-collapsible mw-collapsed">
====Magnetic field of a wire====
<div class="mw-collapsible-content">
*[[Magnetic Field of a Long Straight Wire]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Magnetic field of a current-carrying loop====
<div class="mw-collapsible-content">
*[[Magnetic Field of a Loop]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Magnetic field of a Charged Disk====
<div class="mw-collapsible-content">
*[[Magnetic Field of a Disk]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Magnetic dipoles====
<div class="mw-collapsible-content">
*[[Magnetic Dipole Moment]]
*[[Bar Magnet]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Atomic structure of magnets====
<div class="mw-collapsible-content">
*[[Atomic Structure of Magnets]]
</div>
</div>

===Week 8===
<div class="toccolours mw-collapsible mw-collapsed">
====Steady state current====
<div class="mw-collapsible-content">
*[[Steady State]]
*[[Non Steady State]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Node rule====
<div class="mw-collapsible-content">
*[[Node Rule]]
**In an electric circuit in series, electrons flow from the negative end of a power source, creating a constant current. This current remains consistent at each point in the circuit in series. Sometimes, a circuit is not simply one constant path and may include parts that are in parallel, where the current must travel down two paths such as this:
**[[File:noderule.jpg]]
**In this case, when the current enters a portion of the circuit where the items are in parallel, the total amount of current in must equal the total amount of current out. Therefore, the currents in each branch of the parallel portion must sum up to the amount of current at any other point in series in the circuit.
**For the previous image, the node rules can be written as I_total = I_1 + I_2 and I_total = I_3 + I_4. It is also true that I_1 + I_2 = I_3 + I_4.
**However, each of these currents are different because each point has a different resistance. The current is different for each because it is equal to V/R, and in a parallel circuit, the voltage drop across each point is equal.
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">

====Electric fields and energy in circuits====
<div class="mw-collapsible-content">
*[[Series circuit]]
*[[Node Rule]]
*[[Loop Rule]]
*[[Electric Potential Difference]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Macroscopic analysis of circuits====
<div class="mw-collapsible-content">
*[[Series Circuits]]
*[[Parallel CIrcuits]]
*[[Parallel Circuits vs. Series Circuits*]]
*[[Loop Rule]]
*[[Node Rule]]
*[[Fundamentals of Resistance]]
</div>
</div>

===Week 9===
<div class="toccolours mw-collapsible mw-collapsed">
====Electric field and potential in circuits with capacitors====
<div class="mw-collapsible-content">
*[[Charging and Discharging a Capacitor]]
*[[RC Circuit]]
*[[R Circuit]]
*[[AC and DC]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Magnetic forces on charges and currents====
<div class="mw-collapsible-content">
*[[Magnetic Force]]
*[[Lorentz Force]]
*[[Applying Magnetic Force to Currents]]
*[[Magnetic Force in a Moving Reference Frame]]
*[[Right-Hand Rule]]
*[[Analysis of Railgun vs Coil gun technologies]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Electric and magnetic forces====
<div class="mw-collapsible-content">
*[[Electric Force]]
*[[Magnetic Force]]
*[[Lorentz Force]]
*[[VPython Modelling of Electric and Magnetic Forces]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Velocity selector====
<div class="mw-collapsible-content">
*[[Lorentz Force]]
*[[Combining Electric and Magnetic Forces]]
</div>
</div>

===Week 10===
<div class="toccolours mw-collapsible mw-collapsed">
====The Hall effect====
<div class="mw-collapsible-content">
*[[Hall Effect]]
*[[Right-Hand Rule]]
*[[Motional Emf]]
*[[Magnetic Force]]
*[[Magnetic Torque]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Motional EMF====
<div class="mw-collapsible-content">
*[[Motional Emf]]
*[[Motional Emf using Faraday's Law]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Magnetic force====
<div class="mw-collapsible-content">
*[[Magnetic Force]]
*[[Lorentz Force]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Magnetic torque====
<div class="mw-collapsible-content">
*[[Magnetic Torque]]
*[[Right-Hand Rule]]
</div>
</div>

===Week 12===

<div class="toccolours mw-collapsible mw-collapsed">
====Gauss's Law====
<div class="mw-collapsible-content">
*[[Gauss's Flux Theorem]]
*[[Gauss's Law]]
*[[Magnetic Flux]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Ampere's Law====
<div class="mw-collapsible-content">
*[[Ampere's Law]]
*[[Ampere-Maxwell Law]]
*[[Magnetic Field of Coaxial Cable Using Ampere's Law]]
*[[Magnetic Field of a Long Thick Wire Using Ampere's Law]]
*[[Magnetic Field of a Toroid Using Ampere's Law]]
*[[Magnetic Field of a Solenoid Using Ampere's Law]]
*[[The Differential Form of Ampere's Law]]
</div>
</div>

===Week 13===
<div class="toccolours mw-collapsible mw-collapsed">
====Semiconductors====
<div class="mw-collapsible-content">
*[[Semiconductor Devices]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Faraday's Law====
<div class="mw-collapsible-content">
*[[Faraday's Law]]
*[[Motional Emf using Faraday's Law]]
*[[Lenz's Law]]

</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">

====Maxwell's equations====
<div class="mw-collapsible-content">
*[[Gauss's Law]]
*[[Magnetic Flux]]
*[[Ampere's Law]]
*[[Faraday's Law]]
*[[Maxwell's Electromagnetic Theory]]
</div>
</div>

===Week 14===
<div class="toccolours mw-collapsible mw-collapsed">
====Circuits revisited====
<div class="mw-collapsible-content">

*[[RLC Circuits]]
*[[LR Circuits]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">

====Inductors====
<div class="mw-collapsible-content">
*[[Inductors]]
*[[Current in an LC Circuit]]
*[[Current in an RL Circuit]]
</div>
</div>

===Week 15===
<div class="toccolours mw-collapsible mw-collapsed">
==== Electromagnetic Radiation ====
<div class="mw-collapsible-content">
*[[Electromagnetic Radiation]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Sparks in the air====
<div class="mw-collapsible-content">
*[[Sparks in Air]]
*[[Spark Plugs]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Superconductors====
<div class="mw-collapsible-content">
*[[Superconducters]]
*[[Superconductors]]
*[[Meissner effect]]
</div>
</div>
</div>

<div style="float:left; width:30%; padding:1%;">

==Physics 3==

===Week 1===
<div class="toccolours mw-collapsible mw-collapsed">
====Classical Physics====
<div class="mw-collapsible-content">
</div>
</div>

===Week 2===
<div class="toccolours mw-collapsible mw-collapsed">
====Special Relativity====
<div class="mw-collapsible-content">
*[[Frame of Reference]]
*[[Einstein's Theory of Special Relativity]]
*[[Time Dilation]]
*[[Einstein's Theory of General Relativity]]
*[[Albert A. Micheleson & Edward W. Morley]]
*[[Magnetic Force in a Moving Reference Frame]]
</div>
</div>

===Week 3===
<div class="toccolours mw-collapsible mw-collapsed">
====Photons====
<div class="mw-collapsible-content">
*[[Spontaneous Photon Emission]]
*[[Light Scattering: Why is the Sky Blue]]
*[[Lasers]]
*[[Electronic Energy Levels and Photons]]
*[[Quantum Properties of Light]]
</div>
</div>

===Week 4===
<div class="toccolours mw-collapsible mw-collapsed">
====Matter Waves====
<div class="mw-collapsible-content">
*[[Wave-Particle Duality]]
</div>
</div>

===Week 5===
<div class="toccolours mw-collapsible mw-collapsed">
====Wave Mechanics====
<div class="mw-collapsible-content">
*[[Standing Waves]]
*[[Wavelength]]
*[[Wavelength and Frequency]]
*[[Mechanical Waves]]
*[[Transverse and Longitudinal Waves]]
</div>
</div>

===Week 6===
<div class="toccolours mw-collapsible mw-collapsed">
====Rutherford-Bohr Model====
<div class="mw-collapsible-content">
*[[Rutherford Experiment and Atomic Collisions]]
*[[Bohr Model]]
*[[Quantized energy levels]]
*[[Energy graphs and the Bohr model]]
</div>
</div>

===Week 7===
<div class="toccolours mw-collapsible mw-collapsed">
====The Hydrogen Atom====
<div class="mw-collapsible-content">
*[[Quantum Theory]]
*[[Atomic Theory]]
</div>
</div>

===Week 8===
<div class="toccolours mw-collapsible mw-collapsed">
====Many-Electron Atoms====
<div class="mw-collapsible-content">
*[[Quantum Theory]]
*[[Atomic Theory]]
*[[Pauli exclusion principle]]
</div>
</div>

===Week 9===
<div class="toccolours mw-collapsible mw-collapsed">
====Molecules====
<div class="mw-collapsible-content">
</div>
</div>

===Week 10===
<div class="toccolours mw-collapsible mw-collapsed">
====Statistical Physics====
<div class="mw-collapsible-content">
</div>
</div>

===Week 11===
<div class="toccolours mw-collapsible mw-collapsed">
====Condensed Matter Physics====
<div class="mw-collapsible-content">
</div>
</div>

===Week 12===
<div class="toccolours mw-collapsible mw-collapsed">
====The Nucleus====
<div class="mw-collapsible-content">
</div>
</div>

===Week 13===
<div class="toccolours mw-collapsible mw-collapsed">
====Nuclear Physics====
<div class="mw-collapsible-content">
*[[Nuclear Fission]]
*[[Nuclear Energy from Fission and Fusion]]
</div>
</div>

===Week 14===
<div class="toccolours mw-collapsible mw-collapsed">
====Particle Physics====
<div class="mw-collapsible-content">
*[[Elementary Particles and Particle Physics Theory]]
*[[String Theory]]
</div>
</div>
</div>

Potential Difference Path Independence, claimed by Aditya Mohile

2016-11-28T01:54:30Z

Zwright8: /* References */

''Claimed by Aditya Mohile (Spring 2016)''

'''Claimed by Zachary Wright (Fall 2016)'''

Potential difference or Voltage is a scalar quantity that measures the difference of electric potential energy between two points. Potential energy differences depend only on final and initial states and is independent of the path taken to arrive at those states. Likewise, the potential difference between a final and initial state is independent of the path taken to arrive at those states, therefore the potential difference is said to be path independent. The units of measurement for potential difference is joules per coulomb, which is equal to volts.
==The Main Idea==

The potential difference between two points does not depend on the path taken to get from one point to another because the difference between the locations of the two points remains the same regardless of the path taken to get from one point to another.

===A Mathematical Model===
:<math>\Delta V_{AB} = \int_{A}^{B} \vec{E} \cdot d\vec{l} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>

The potential difference, <math>\Delta V_{AB}</math>, is the integral of the electrical field from point A to point B. This integral, however, is the dot product of <math>\vec{E}</math> and <math>d\vec{l}</math> which is equal to the sum of the negative electric field multiplied by the difference of the x,y and z components of the two points, as see from the equation above. This proves that the path taken from point A to point B does not matter, only the difference in the positions of points A and B are important in calculating the potential difference.

[[File:Electrostatic_definition_of_voltage.jpg]]

As shown above, the potential difference between points A and B in the electric field E is independent of the path.

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===
[[File:Wiki_simple.jpg|center]]
'''Problem''': Locations A, B, and C are in a region of uniform electric field, as shown in the diagram above. Location A is at < -0.6, 0, 0> m. Location B is at < 0.4, 0, 0>. Location C is at < 0.4, -0.3, 0>m. In the region the electric field = < -750, 400, 0> N/C. Calculate the potential difference between A and C traveling from A to B then to C. Calculate the potential difference traveling from A to C directly.

'''Step 1'''
:Calculate the displacement vector from A to B.
:<math>\Delta l = < 0.4, 0, 0> - < -0.6, 0, 0> = < 1, 0, 0></math>
'''Step 2'''
:Calculate the potential difference between A and B.
:<math>\Delta V_{AB} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AB} = -(-750*1 + 400*0 + 0*0) = 750V</math>
'''Step 3'''
:Calculate the displacement vector from B to C.
:<math>\Delta l = < 0.4, -0.3, 0> - < 0.4, 0, 0> = < 0, -0.3, 0></math>
'''Step 4'''
:Calculate the potential difference between B and C.
:<math>\Delta V_{BC} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{BC} = -(-750*0 + 400*-0.3 + 0*0) = 120V</math>
'''Step 5'''
:Add <math>\Delta V_{AB}</math> to <math>\Delta V_{BC}</math> to get <math>\Delta V_{AC}</math>
:<math>\Delta V_{AC} = \Delta V_{AB} + \Delta V_{BC} = 750 + 120 = 870V</math>
Now you have to calculate the potential difference from A to C directly.

'''Step 1'''
:Calculate the displacement vector from A to C.
:<math>\Delta l = < 0.4, -0.3, 0> - < -0.6, 0, 0> = < 1, -0.3, 0></math>
'''Step 2'''
:Calculate the potential difference between A and C.
:<math>\Delta V_{AB} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AB} = -(-750*1 + 400*-0.3 + 0*0) = 750 + 120 = 870V</math>
As you can see, potential difference from A to C directly is the same as the potential difference from A to B then to C.

===Middling===
[[File:Wiki_middling.jpg|350px|center]]
'''Problem''': A capacitor consists of two charged disks of radius 4.7 m separated by a distance s = 2 mm (see the figure). The magnitude of the charge on each disk is 56 µC. Consider points A, B, C, and D inside the capacitor, as shown in the diagram. The distance s1 = 1.5 mm, and the distance s2 = 0.6 mm. (Assume the +x axis is to the right, the +y axis is up, and the +z axis is out.) Find the potential difference between B and D, traveling through A. Find the potential difference from B to D directly.
'''Step 1'''
:Calculate the electric field inside the capacitor.
:<math>\vec E_{cap} = \frac{Q/A}{\epsilon_{0}}</math>
:<math>\vec E_{cap} = \frac{(56*10^{-6})/(\pi*4.7^{2})}{8.85*10^{-12}} = < 91179.912, 0, 0></math>
The electric field only has an x-component because the field travels from the positively charged plate to the negatively charged plate, traveling in the +x direction.

'''Step 2'''
:Calculate the displacement vector from B to A.
:Set the origin at B.
:<math>\Delta l = < s_{1}, 0, 0> - < 0, 0, 0> = < s_{1}, 0, 0> = < 0.0015, 0, 0></math>
'''Step 3'''
:Calculate the potential difference between B and A.
:<math>\Delta V_{BA} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{BA} = -(91179.912*0.0015 + 0*0 + 0*0) = -136.770V</math>
'''Step 4'''
:Calculate the displacement vector from A to D
:<math>\Delta l = < s_{1}, s_{2}, 0> - < s_{1}, 0, 0> = < 0, s_{2}, 0> = < 0, s_{2}, 0> = < 0, 0.006, 0></math>
'''Step 5'''
:Calculate the potential difference between A and D.
:<math>\Delta V_{AD} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AD} = -(91179.912*0 + 0*0 + 0*0) = 0V</math>
'''Step 6'''
:Add <math>\Delta V_{BA}</math> to <math>\Delta V_{AD}</math> to get <math>\Delta V_{BD}</math>
:<math>\Delta V_{BD} = \Delta V_{BA} + \Delta V_{AD} = -136.770 + 0 = -136.770V</math>
Notice that the voltage is negative since we are going from the positively charged plate (high potential) to the negatively charged plate (low potential) losing potential energy in the process.

Now you have to calculate the potential difference from A to C directly.

'''Step 1'''
:Calculate the displacement vector from B to D.
:<math>\Delta l = < s_{1}, s_{2}, 0> - < 0, 0, 0> = < s_{1}, s_{2}, 0> = < s_{1}, s_{2}, 0> = < 0.0015, 0.006, 0></math>
'''Step 2'''
:Calculate the potential difference between B and D.
:<math>\Delta V_{AD} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AD} = -(91179.912*0.0015 + 0*0 + 0*0) = -136.770V</math>
Notice that potential difference traveling from B to A then D is the same as the potential difference traveling from B to D directly. This exemplifies that the potential difference does not depend on the path taken to get from one point to another.

===Difficult===
[[File:Wiki_difficult.png|center]]
'''Problem''': A uniform spherical shell of charge +Q is centered at point B, as shown in the figure below. Show that ΔV = VC − VA is independent of path by calculating ΔV for each of these two paths (actually do the integrals). (Use the following as necessary: Q, R for the radius of the spherical shell, r1 for the radius of the circular arc A → D → C with B at its center, and ε0.)

'''Step 1'''
:Calculate the potential difference from A to C following the path A → B → C.
:First, calculate the potential difference from A to B.
:<math>\Delta V_{AB} = \int_{A}^{B} \vec{E} \cdot d\vec{l}</math>
:In order to calculate the potential difference from A to B, calculate the electric field of the sphere.
:<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}}</math>
:Now plug the electric field calculated above into the integral and solve.
:<math>\Delta V_{AB} = \int_{A}^{B} \vec{E} \cdot d\vec{r} = \int_{r_{1}}^{R} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{R^{2}} - \frac{Q}{r_{1}^{2}})</math>
:Now calculate the potential difference from B to C.
:Use the electric field of the sphere calculated above to solve the below integral.
:<math>\Delta V_{BC} = \int_{B}^{C} \vec{E} \cdot d\vec{l}</math>
:<math>\Delta V_{BC} = \int_{R}^{r_{1}} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{R^{2}})</math>
:Now add <math>\Delta V_{AB}</math> to <math>\Delta V_{BC}</math> to get <math>\Delta V_{AC}</math>
:<math>\Delta V_{AC} = \Delta V_{AB} + \Delta V_{BC} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{R^{2}} - \frac{Q}{r_{1}^{2}}) + \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{R^{2}}) = 0V </math>
'''Step 2'''
:Calculate the potential difference from A to C following the path A → D → C.
:First calculate the potential difference from A to D.
:<math>\Delta V_{AD} = \int_{A}^{D} \vec{E} \cdot d\vec{l}</math>
:Use the same electric field of the sphere calculated in step 1, shown below.
:<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}}</math>
:Now plug the electric field calculated above into the integral and solve.
:<math>\Delta V_{AD} = \int_{A}^{D} \vec{E} \cdot d\vec{r} = \int_{r_{1}}^{r_{1}} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}}) = 0V</math>
:Now calculate the potential difference from D to C.
:Use the electric field of the sphere calculated above to solve the below integral.
:<math>\Delta V_{DC} = \int_{D}^{C} \vec{E} \cdot d\vec{l}</math>
:<math>\Delta V_{DC} = \int_{r_{1}}^{r_{1}} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}})</math>
:Now add <math>\Delta V_{AD}</math> to <math>\Delta V_{DC}</math> to get <math>\Delta V_{AC}</math>
:<math>\Delta V_{AC} = \Delta V_{AD} + \Delta V_{DC} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}}) + \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}}) = 0V </math>
:Notice that either path taken from A to C results in the same potential difference of 0V.

==Connectedness==
#How is this topic connected to something that you are interested in?
::I am very interested in electronics and circuitry so knowing that potential difference is path independent definitely helps in determining the voltage between 2 points in a complicated circuit with many paths.
#How is it connected to your major?
::I am an electrical engineering major so this knowledge is very applicable to determining the voltage between 2 points in a circuit that contains many loops or is very complex.
#In Circuits
::Path Independence of potential difference is critical in solving circuit equations. Many values in a circuit can be calculated using these principles and the Kirchoff's Voltage Law.

==History==

Path independence of potential difference stems from the Law of Conservation of Energy. Many noteworthy scientists worked to affirm this law such as: Isaac Newton, Johann and Daniel Bernoulli, Gottfried Leibniz, and many others.

The unit of measurement for potential difference is joules per coulomb, which is equal to a volt. The unit Volt is named in honor of the Italian scientist Alessandro Volta, who was an innovator in the fields of electricity and power.

Another notable Italian scientist,Luigi Galvani, discovered something he called "animal electricity" which was two different metals connected in series with a frog leg and each other. However, Alessandro Volta, another Italian physicist, realized that the frog leg was acting as a conductor and a detector of electricity, or an electrolyte. He later replaced the frog leg with brine soaked paper and noticed the same results. Through this, Volta discovered electromotive force relating to a galvanic cell, realizing that the difference between the two electrode potentials caused electricity to flow.

== See also ==

https://en.wikipedia.org/wiki/Voltage

===Further reading===

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elewor.html#c2

===External links===

http://mathwiki.ucdavis.edu/Core/Calculus/Vector_Calculus/Integration_in_Vector_Fields/Path_Independence,_Conservative_Fields,_and_Potential_Functions

==References==

Matter & Interactions volume II by Ruth W. Chabay and Bruce A. Sherwood.

[[Category:Energy]]

Potential Difference Path Independence, claimed by Aditya Mohile

2016-11-28T01:50:54Z

Zwright8: /* See also */

Potential Difference Path Independence, claimed by Aditya Mohile

2016-11-28T01:50:23Z

Zwright8: /* See also */

''Claimed by Aditya Mohile (Spring 2016)''

'''Claimed by Zachary Wright (Fall 2016)'''

Potential difference or Voltage is a scalar quantity that measures the difference of electric potential energy between two points. Potential energy differences depend only on final and initial states and is independent of the path taken to arrive at those states. Likewise, the potential difference between a final and initial state is independent of the path taken to arrive at those states, therefore the potential difference is said to be path independent. The units of measurement for potential difference is joules per coulomb, which is equal to volts.
==The Main Idea==

The potential difference between two points does not depend on the path taken to get from one point to another because the difference between the locations of the two points remains the same regardless of the path taken to get from one point to another.

===A Mathematical Model===
:<math>\Delta V_{AB} = \int_{A}^{B} \vec{E} \cdot d\vec{l} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>

The potential difference, <math>\Delta V_{AB}</math>, is the integral of the electrical field from point A to point B. This integral, however, is the dot product of <math>\vec{E}</math> and <math>d\vec{l}</math> which is equal to the sum of the negative electric field multiplied by the difference of the x,y and z components of the two points, as see from the equation above. This proves that the path taken from point A to point B does not matter, only the difference in the positions of points A and B are important in calculating the potential difference.

[[File:Electrostatic_definition_of_voltage.jpg]]

As shown above, the potential difference between points A and B in the electric field E is independent of the path.

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===
[[File:Wiki_simple.jpg|center]]
'''Problem''': Locations A, B, and C are in a region of uniform electric field, as shown in the diagram above. Location A is at < -0.6, 0, 0> m. Location B is at < 0.4, 0, 0>. Location C is at < 0.4, -0.3, 0>m. In the region the electric field = < -750, 400, 0> N/C. Calculate the potential difference between A and C traveling from A to B then to C. Calculate the potential difference traveling from A to C directly.

'''Step 1'''
:Calculate the displacement vector from A to B.
:<math>\Delta l = < 0.4, 0, 0> - < -0.6, 0, 0> = < 1, 0, 0></math>
'''Step 2'''
:Calculate the potential difference between A and B.
:<math>\Delta V_{AB} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AB} = -(-750*1 + 400*0 + 0*0) = 750V</math>
'''Step 3'''
:Calculate the displacement vector from B to C.
:<math>\Delta l = < 0.4, -0.3, 0> - < 0.4, 0, 0> = < 0, -0.3, 0></math>
'''Step 4'''
:Calculate the potential difference between B and C.
:<math>\Delta V_{BC} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{BC} = -(-750*0 + 400*-0.3 + 0*0) = 120V</math>
'''Step 5'''
:Add <math>\Delta V_{AB}</math> to <math>\Delta V_{BC}</math> to get <math>\Delta V_{AC}</math>
:<math>\Delta V_{AC} = \Delta V_{AB} + \Delta V_{BC} = 750 + 120 = 870V</math>
Now you have to calculate the potential difference from A to C directly.

'''Step 1'''
:Calculate the displacement vector from A to C.
:<math>\Delta l = < 0.4, -0.3, 0> - < -0.6, 0, 0> = < 1, -0.3, 0></math>
'''Step 2'''
:Calculate the potential difference between A and C.
:<math>\Delta V_{AB} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AB} = -(-750*1 + 400*-0.3 + 0*0) = 750 + 120 = 870V</math>
As you can see, potential difference from A to C directly is the same as the potential difference from A to B then to C.

===Middling===
[[File:Wiki_middling.jpg|350px|center]]
'''Problem''': A capacitor consists of two charged disks of radius 4.7 m separated by a distance s = 2 mm (see the figure). The magnitude of the charge on each disk is 56 µC. Consider points A, B, C, and D inside the capacitor, as shown in the diagram. The distance s1 = 1.5 mm, and the distance s2 = 0.6 mm. (Assume the +x axis is to the right, the +y axis is up, and the +z axis is out.) Find the potential difference between B and D, traveling through A. Find the potential difference from B to D directly.
'''Step 1'''
:Calculate the electric field inside the capacitor.
:<math>\vec E_{cap} = \frac{Q/A}{\epsilon_{0}}</math>
:<math>\vec E_{cap} = \frac{(56*10^{-6})/(\pi*4.7^{2})}{8.85*10^{-12}} = < 91179.912, 0, 0></math>
The electric field only has an x-component because the field travels from the positively charged plate to the negatively charged plate, traveling in the +x direction.

'''Step 2'''
:Calculate the displacement vector from B to A.
:Set the origin at B.
:<math>\Delta l = < s_{1}, 0, 0> - < 0, 0, 0> = < s_{1}, 0, 0> = < 0.0015, 0, 0></math>
'''Step 3'''
:Calculate the potential difference between B and A.
:<math>\Delta V_{BA} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{BA} = -(91179.912*0.0015 + 0*0 + 0*0) = -136.770V</math>
'''Step 4'''
:Calculate the displacement vector from A to D
:<math>\Delta l = < s_{1}, s_{2}, 0> - < s_{1}, 0, 0> = < 0, s_{2}, 0> = < 0, s_{2}, 0> = < 0, 0.006, 0></math>
'''Step 5'''
:Calculate the potential difference between A and D.
:<math>\Delta V_{AD} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AD} = -(91179.912*0 + 0*0 + 0*0) = 0V</math>
'''Step 6'''
:Add <math>\Delta V_{BA}</math> to <math>\Delta V_{AD}</math> to get <math>\Delta V_{BD}</math>
:<math>\Delta V_{BD} = \Delta V_{BA} + \Delta V_{AD} = -136.770 + 0 = -136.770V</math>
Notice that the voltage is negative since we are going from the positively charged plate (high potential) to the negatively charged plate (low potential) losing potential energy in the process.

Now you have to calculate the potential difference from A to C directly.

'''Step 1'''
:Calculate the displacement vector from B to D.
:<math>\Delta l = < s_{1}, s_{2}, 0> - < 0, 0, 0> = < s_{1}, s_{2}, 0> = < s_{1}, s_{2}, 0> = < 0.0015, 0.006, 0></math>
'''Step 2'''
:Calculate the potential difference between B and D.
:<math>\Delta V_{AD} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AD} = -(91179.912*0.0015 + 0*0 + 0*0) = -136.770V</math>
Notice that potential difference traveling from B to A then D is the same as the potential difference traveling from B to D directly. This exemplifies that the potential difference does not depend on the path taken to get from one point to another.

===Difficult===
[[File:Wiki_difficult.png|center]]
'''Problem''': A uniform spherical shell of charge +Q is centered at point B, as shown in the figure below. Show that ΔV = VC − VA is independent of path by calculating ΔV for each of these two paths (actually do the integrals). (Use the following as necessary: Q, R for the radius of the spherical shell, r1 for the radius of the circular arc A → D → C with B at its center, and ε0.)

'''Step 1'''
:Calculate the potential difference from A to C following the path A → B → C.
:First, calculate the potential difference from A to B.
:<math>\Delta V_{AB} = \int_{A}^{B} \vec{E} \cdot d\vec{l}</math>
:In order to calculate the potential difference from A to B, calculate the electric field of the sphere.
:<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}}</math>
:Now plug the electric field calculated above into the integral and solve.
:<math>\Delta V_{AB} = \int_{A}^{B} \vec{E} \cdot d\vec{r} = \int_{r_{1}}^{R} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{R^{2}} - \frac{Q}{r_{1}^{2}})</math>
:Now calculate the potential difference from B to C.
:Use the electric field of the sphere calculated above to solve the below integral.
:<math>\Delta V_{BC} = \int_{B}^{C} \vec{E} \cdot d\vec{l}</math>
:<math>\Delta V_{BC} = \int_{R}^{r_{1}} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{R^{2}})</math>
:Now add <math>\Delta V_{AB}</math> to <math>\Delta V_{BC}</math> to get <math>\Delta V_{AC}</math>
:<math>\Delta V_{AC} = \Delta V_{AB} + \Delta V_{BC} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{R^{2}} - \frac{Q}{r_{1}^{2}}) + \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{R^{2}}) = 0V </math>
'''Step 2'''
:Calculate the potential difference from A to C following the path A → D → C.
:First calculate the potential difference from A to D.
:<math>\Delta V_{AD} = \int_{A}^{D} \vec{E} \cdot d\vec{l}</math>
:Use the same electric field of the sphere calculated in step 1, shown below.
:<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}}</math>
:Now plug the electric field calculated above into the integral and solve.
:<math>\Delta V_{AD} = \int_{A}^{D} \vec{E} \cdot d\vec{r} = \int_{r_{1}}^{r_{1}} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}}) = 0V</math>
:Now calculate the potential difference from D to C.
:Use the electric field of the sphere calculated above to solve the below integral.
:<math>\Delta V_{DC} = \int_{D}^{C} \vec{E} \cdot d\vec{l}</math>
:<math>\Delta V_{DC} = \int_{r_{1}}^{r_{1}} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}})</math>
:Now add <math>\Delta V_{AD}</math> to <math>\Delta V_{DC}</math> to get <math>\Delta V_{AC}</math>
:<math>\Delta V_{AC} = \Delta V_{AD} + \Delta V_{DC} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}}) + \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}}) = 0V </math>
:Notice that either path taken from A to C results in the same potential difference of 0V.

==Connectedness==
#How is this topic connected to something that you are interested in?
::I am very interested in electronics and circuitry so knowing that potential difference is path independent definitely helps in determining the voltage between 2 points in a complicated circuit with many paths.
#How is it connected to your major?
::I am an electrical engineering major so this knowledge is very applicable to determining the voltage between 2 points in a circuit that contains many loops or is very complex.
#In Circuits
::Path Independence of potential difference is critical in solving circuit equations. Many values in a circuit can be calculated using these principles and the Kirchoff's Voltage Law.

==History==

Path independence of potential difference stems from the Law of Conservation of Energy. Many noteworthy scientists worked to affirm this law such as: Isaac Newton, Johann and Daniel Bernoulli, Gottfried Leibniz, and many others.

The unit of measurement for potential difference is joules per coulomb, which is equal to a volt. The unit Volt is named in honor of the Italian scientist Alessandro Volta, who was an innovator in the fields of electricity and power.

Another notable Italian scientist,Luigi Galvani, discovered something he called "animal electricity" which was two different metals connected in series with a frog leg and each other. However, Alessandro Volta, another Italian physicist, realized that the frog leg was acting as a conductor and a detector of electricity, or an electrolyte. He later replaced the frog leg with brine soaked paper and noticed the same results. Through this, Volta discovered electromotive force relating to a galvanic cell, realizing that the difference between the two electrode potentials caused electricity to flow.

== See also ==

https://en.wikipedia.org/wiki/Voltage

https://en.wikipedia.org/wiki/Conservation_of_energy

===Further reading===

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elewor.html#c2

===External links===

http://mathwiki.ucdavis.edu/Core/Calculus/Vector_Calculus/Integration_in_Vector_Fields/Path_Independence,_Conservative_Fields,_and_Potential_Functions

==References==

Matter & Interactions volume II by Ruth W. Chabay and Bruce A. Sherwood.

[[Category:Energy]]

Main Page

2016-11-28T01:48:20Z

Zwright8: /* Path independence and round trip potential */

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<div style="float:left; width:30%; padding:1%;">
==Physics 1==
===Week 1===
<div class="toccolours mw-collapsible mw-collapsed">
====Help with VPython====
<div class="mw-collapsible-content">
*[[VPython]]
*[[VPython basics]]
*[[VPython Common Errors and Troubleshooting]]
*[[VPython Functions]]
*[[VPython Lists]]
*[[VPython Loops]]
*[[VPython Multithreading]]
*[[VPython Animation]]
*[[VPython Objects]]
*[[VPython 3D Objects]]
*[[VPython Reference]]
*[[VPython MapReduceFilter]]
*[[VPython GUIs]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">

====Vectors and Units====
<div class="mw-collapsible-content">
*[[Vectors]]
*[[SI Units]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">

====Interactions====
<div class="mw-collapsible-content">
*[[Types of Interactions and How to Detect Them]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Velocity and Momentum====
<div class="mw-collapsible-content">
*[[Newton's First Law of Motion]]
*[[Velocity]]
*[[Mass]]
*[[Speed and Velocity]]
*[[Relative Velocity]]
*[[Derivation of Average Velocity]]
*[[2-Dimensional Motion]]
*[[3-Dimensional Position and Motion]]
</div>
</div>

===Week 2===
<div class="toccolours mw-collapsible mw-collapsed">
====Momentum and the Momentum Principle====
<div class="mw-collapsible-content">
*[[Momentum Principle]]
*[[Inertia]]
*[[Net Force]]
*[[Derivation of the Momentum Principle]]
*[[Impulse Momentum]]
*[[Acceleration]]
*[[Momentum with respect to external Forces]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Iterative Prediction with a Constant Force====
<div class="mw-collapsible-content">
*[[Newton’s Second Law of Motion]]
*[[Iterative Prediction]]
*[[Kinematics]]
*[[Newton’s Laws and Linear Momentum]]
*[[Projectile Motion]]
</div>
</div>

===Week 3===
<div class="toccolours mw-collapsible mw-collapsed">
====Analytic Prediction with a Constant Force====
<div class="mw-collapsible-content">
*[[Analytical Prediction]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Iterative Prediction with a Varying Force====
<div class="mw-collapsible-content">
*[[Predicting Change in multiple dimensions]]
*[[Spring Force]]
*[[Hooke's Law]]
*[[Simple Harmonic Motion]]
*[[Iterative Prediction of Spring-Mass System]]
*[[Terminal Speed]]
*[[Determinism]]
</div>
</div>

===Week 4===
<div class="toccolours mw-collapsible mw-collapsed">
====Fundamental Interactions====
<div class="mw-collapsible-content">
*[[Gravitational Force]]
*[[An Application of Gravitational Potential]]
*[[Electric Force]]
*[[Reciprocity]]
</div>
</div>

===Week 5===
<div class="toccolours mw-collapsible mw-collapsed">
====Conservation of Momentum====
<div class="mw-collapsible-content">
*[[Conservation of Momentum]]
</div>
</div>
<div class="toccolours mw-collapsible mw-collapsed">

====Properties of Matter====
<div class="mw-collapsible-content">
*[[Kinds of Matter]]
*[[Ball and Spring Model of Matter]]
*[[Density]]
*[[Length and Stiffness of an Interatomic Bond]]
*[[Young's Modulus]]
*[[Speed of Sound in Solids]]
*[[Malleability]]
*[[Ductility]]
*[[Weight]]
*[[Hardness]]
*[[Boiling Point]]
*[[Melting Point]]
</div>
</div>

===Week 6===
<div class="toccolours mw-collapsible mw-collapsed">
====Identifying Forces====
<div class="mw-collapsible-content">
*[[Free Body Diagram]]
*[[Inclined Plane]]
*[[Compression or Normal Force]]
*[[Tension]]
</div>
</div>
<div class="toccolours mw-collapsible mw-collapsed">

====Curving Motion====
<div class="mw-collapsible-content">
*[[Curving Motion]]
*[[Centripetal Force and Curving Motion]]
*[[Perpetual Freefall (Orbit)]]
</div>
</div>

===Week 7===
<div class="toccolours mw-collapsible mw-collapsed">
====Energy Principle====
<div class="mw-collapsible-content">
*[[The Energy Principle]]
*[[Conservation of Energy]]
*[[Kinetic Energy]]
*[[Work]]
*[[Power (Mechanical)]]
</div>
</div>

===Week 8===
<div class="toccolours mw-collapsible mw-collapsed">
====Work by Non-Constant Forces====
<div class="mw-collapsible-content">
*[[Work Done By A Nonconstant Force]]
</div>
</div>
<div class="toccolours mw-collapsible mw-collapsed">
====Potential Energy====
<div class="mw-collapsible-content">
*[[Potential Energy]]
*[[Potential Energy of Macroscopic Springs]]
*[[Spring Potential Energy]]
*[[Ball and Spring Model]]
*[[Gravitational Potential Energy]]
*[[Energy Graphs]]
*[[Escape Velocity]]
</div>
</div>

===Week 9===
<div class="toccolours mw-collapsible mw-collapsed">
====Multiparticle Systems====
<div class="mw-collapsible-content">
*[[Center of Mass]]
*[[Multi-particle analysis of Momentum]]
*[[Momentum with respect to external Forces]]
*[[Potential Energy of a Multiparticle System]]
*[[Work and Energy for an Extended System]]
*[[Internal Energy]]
**[[Potential Energy of a Pair of Neutral Atoms]]
</div>
</div>

===Week 10===
<div class="toccolours mw-collapsible mw-collapsed">
====Choice of System====
<div class="mw-collapsible-content">
*[[System & Surroundings]]
</div>
</div>
<div class="toccolours mw-collapsible mw-collapsed">
====Thermal Energy, Dissipation and Transfer of Energy====
<div class="mw-collapsible-content">
*[[Thermal Energy]]
*[[Specific Heat]]
*[[Heat Capacity]]
*[[Specific Heat Capacity]]
*[[First Law of Thermodynamics]]
*[[Second Law of Thermodynamics and Entropy]]
*[[Temperature]]
*[[Predicting Change]]
*[[Energy Transfer due to a Temperature Difference]]
*[[Transformation of Energy]]
*[[The Maxwell-Boltzmann Distribution]]
*[[Air Resistance]]
</div>
</div>
<div class="toccolours mw-collapsible mw-collapsed">
====Rotational and Vibrational Energy====
<div class="mw-collapsible-content">
*[[Translational, Rotational and Vibrational Energy]]
</div>
</div>

===Week 11===
<div class="toccolours mw-collapsible mw-collapsed">
====Different Models of a System====
<div class="mw-collapsible-content">
*[[Point Particle Systems]]
*[[Real Systems]]
</div>
</div>
<div class="toccolours mw-collapsible mw-collapsed">
====Models of Friction====
<div class="mw-collapsible-content">
*[[Friction]]
*[[Static Friction]]
</div>
</div>

===Week 12===
<div class="toccolours mw-collapsible mw-collapsed">
====Collisions====
<div class="mw-collapsible-content">
*[[Newton's Third Law of Motion]]
*[[Collisions]]
*[[Elastic Collisions]]
*[[Inelastic Collisions]]
*[[Maximally Inelastic Collision]]
*[[Head-on Collision of Equal Masses]]
*[[Head-on Collision of Unequal Masses]]
*[[Scattering: Collisions in 2D and 3D]]
*[[Rutherford Experiment and Atomic Collisions]]
*[[Coefficient of Restitution]]
</div>
</div>

===Week 13===
<div class="toccolours mw-collapsible mw-collapsed">
====Rotations====
<div class="mw-collapsible-content">
*[[Rotation]]
*[[Angular Velocity]]
*[[Eulerian Angles]]
</div>
</div>
<div class="toccolours mw-collapsible mw-collapsed">
====Angular Momentum====
<div class="mw-collapsible-content">
*[[Total Angular Momentum]]
*[[Translational Angular Momentum]]
*[[Rotational Angular Momentum]]
*[[The Angular Momentum Principle]]
*[[Angular Momentum Compared to Linear Momentum]]
*[[Angular Impulse]]
*[[Predicting the Position of a Rotating System]]
*[[Angular Momentum of Multiparticle Systems]]
*[[The Moments of Inertia]]
*[[Moment of Inertia for a cylinder]]
*[[Right Hand Rule]]
</div>
</div>

===Week 14===
<div class="toccolours mw-collapsible mw-collapsed">
====Analyzing Motion with and without Torque====
<div class="mw-collapsible-content">
*[[Torque]]
*[[Torque 2]]
*[[Systems with Zero Torque]]
*[[Systems with Nonzero Torque]]
*[[Torque vs Work]]
*[[Gyroscopes]]
</div>
</div>

===Week 15===
<div class="toccolours mw-collapsible mw-collapsed">
====Introduction to Quantum Concepts====
<div class="mw-collapsible-content">
*[[Bohr Model]]
*[[Energy graphs and the Bohr model]]
*[[Quantized energy levels]]
</div>
</div>
</div>

<div style="float:left; width:30%; padding:1%;">
==Physics 2==
===Week 1===
<div class="toccolours mw-collapsible mw-collapsed">
====3D Vectors====
<div class="mw-collapsible-content">
*[[Vectors]]
*[[Right-Hand Rule]]
*[[Right Hand Rule]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Electric field====
<div class="mw-collapsible-content">
*[[Electric Field]]
*[[Electric Field and Electric Potential]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">

====Electric force====
<div class="mw-collapsible-content">
*[[Electric Force]]
*[[Lorentz Force]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
Aniruddha Nadkarni

====Electric field of a point particle====
<div class="mw-collapsible-content">
*[[Point Charge]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">

====Superposition====
<div class="mw-collapsible-content">
*[[Superposition Principle]]
*[[Superposition principle]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Dipoles====
<div class="mw-collapsible-content">
*[[Electric Dipole]]
*[[Magnetic Dipole]]
</div>
</div>

===Week 2===
<div class="toccolours mw-collapsible mw-collapsed">
====Interactions of charged objects====
<div class="mw-collapsible-content">
*[[Electric Field]]
*[[Electric Potential]]
*[[Electric Force]]
*[[Lorentz Force]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Tape experiments====
<div class="mw-collapsible-content">
*[[Polarization]]
*[[Electric Polarization]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Polarization====
<div class="mw-collapsible-content">
*[[Polarization]]
*[[Electric Polarization]]
*[[Polarization of an Atom]]
</div>
</div>

===Week 3===
<div class="toccolours mw-collapsible mw-collapsed">
====Insulators====
<div class="mw-collapsible-content">
*[[Insulators]]
*[[Potential Difference in an Insulator]]
*[[Charged Conductor and Charged Insulator]]
*[[Charged conductor and charged insulator]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Conductors====
<div class="mw-collapsible-content">
*[[Conductivity]]
*[[Charge Transfer]]
*[[Resistivity]]
*[[Polarization of a conductor]]
*[[Charged Conductor and Charged Insulator]]
*[[Charged conductor and charged insulator]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Charging and discharging====
<div class="mw-collapsible-content">
*[[Charge Transfer]]
*[[Electrostatic Discharge]]
*[[Charged Conductor and Charged Insulator]]
*[[Charged conductor and charged insulator]]
</div>
</div>

===Week 4===
<div class="toccolours mw-collapsible mw-collapsed">
====Field of a charged rod====
<div class="mw-collapsible-content">
*[[Charged Rod]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Field of a charged ring/disk/capacitor====
<div class="mw-collapsible-content">
*[[Charged Ring]]
*[[Charged Disk]]
*[[Charged Capacitor]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Field of a charged sphere====
<div class="mw-collapsible-content">
*[[Charged Spherical Shell]]
*[[Field of a Charged Ball]]
</div>
</div>

===Week 5===
<div class="toccolours mw-collapsible mw-collapsed">
====Potential energy====
<div class="mw-collapsible-content">
*[[Potential Energy - Claimed by Janki Patel]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Electric potential====
<div class="mw-collapsible-content">
*[[Electric Potential]]
*[[Path Independence of Electric Potential]]
*[[Potential Difference Path Independence, claimed by Aditya Mohile]]
*[[Potential Difference in a Uniform Field]]
*[[Potential Difference of Point Charge in a Non-Uniform Field]]
</div>
</div>
<div class="toccolours mw-collapsible mw-collapsed">
====Sign of a potential difference====
<div class="mw-collapsible-content">
*[[Sign of a Potential Difference]]
</div>
</div>
<div class="toccolours mw-collapsible mw-collapsed">
====Potential at a single location====
<div class="mw-collapsible-content">
*[[Electric Potential]]
*[[Potential Difference at One Location]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">

====Path independence and round trip potential====
<div class="mw-collapsible-content">
*[[Path Independence of Electric Potential]]
*[[Potential Difference Path Independence]]
</div>
</div>

===Week 6===
<div class="toccolours mw-collapsible mw-collapsed">
====Electric field and potential in an insulator====
<div class="mw-collapsible-content">
*[[Potential Difference in an Insulator]]
*[[Electric Field in an Insulator]]
</div>
</div>
<div class="toccolours mw-collapsible mw-collapsed">
====Moving charges in a magnetic field====
<div class="mw-collapsible-content">
*[[Magnetic Field]]
*[[Magnetic Force]]
*[[Lorentz Force]]
</div>
</div>
<div class="toccolours mw-collapsible mw-collapsed">
====Biot-Savart Law====
<div class="mw-collapsible-content">
*[[Biot-Savart Law]]
*[[Biot-Savart Law for Currents]]
</div>
</div>
<div class="toccolours mw-collapsible mw-collapsed">
====Moving charges, electron current, and conventional current====
<div class="mw-collapsible-content">
*[[Moving Point Charge]]
*[[Current]]
</div>
</div>

===Week 7===
<div class="toccolours mw-collapsible mw-collapsed">
====Magnetic field of a wire====
<div class="mw-collapsible-content">
*[[Magnetic Field of a Long Straight Wire]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Magnetic field of a current-carrying loop====
<div class="mw-collapsible-content">
*[[Magnetic Field of a Loop]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Magnetic field of a Charged Disk====
<div class="mw-collapsible-content">
*[[Magnetic Field of a Disk]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Magnetic dipoles====
<div class="mw-collapsible-content">
*[[Magnetic Dipole Moment]]
*[[Bar Magnet]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Atomic structure of magnets====
<div class="mw-collapsible-content">
*[[Atomic Structure of Magnets]]
</div>
</div>

===Week 8===
<div class="toccolours mw-collapsible mw-collapsed">
====Steady state current====
<div class="mw-collapsible-content">
*[[Steady State]]
*[[Non Steady State]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Node rule====
<div class="mw-collapsible-content">
*[[Node Rule]]
**In an electric circuit in series, electrons flow from the negative end of a power source, creating a constant current. This current remains consistent at each point in the circuit in series. Sometimes, a circuit is not simply one constant path and may include parts that are in parallel, where the current must travel down two paths such as this:
**[[File:noderule.jpg]]
**In this case, when the current enters a portion of the circuit where the items are in parallel, the total amount of current in must equal the total amount of current out. Therefore, the currents in each branch of the parallel portion must sum up to the amount of current at any other point in series in the circuit.
**For the previous image, the node rules can be written as I_total = I_1 + I_2 and I_total = I_3 + I_4. It is also true that I_1 + I_2 = I_3 + I_4.
**However, each of these currents are different because each point has a different resistance. The current is different for each because it is equal to V/R, and in a parallel circuit, the voltage drop across each point is equal.
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">

====Electric fields and energy in circuits====
<div class="mw-collapsible-content">
*[[Series circuit]]
*[[Node Rule]]
*[[Loop Rule]]
*[[Electric Potential Difference]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Macroscopic analysis of circuits====
<div class="mw-collapsible-content">
*[[Series Circuits]]
*[[Parallel CIrcuits]]
*[[Parallel Circuits vs. Series Circuits*]]
*[[Loop Rule]]
*[[Node Rule]]
*[[Fundamentals of Resistance]]
</div>
</div>

===Week 9===
<div class="toccolours mw-collapsible mw-collapsed">
====Electric field and potential in circuits with capacitors====
<div class="mw-collapsible-content">
*[[Charging and Discharging a Capacitor]]
*[[RC Circuit]]
*[[R Circuit]]
*[[AC and DC]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Magnetic forces on charges and currents====
<div class="mw-collapsible-content">
*[[Magnetic Force]]
*[[Lorentz Force]]
*[[Applying Magnetic Force to Currents]]
*[[Magnetic Force in a Moving Reference Frame]]
*[[Right-Hand Rule]]
*[[Analysis of Railgun vs Coil gun technologies]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Electric and magnetic forces====
<div class="mw-collapsible-content">
*[[Electric Force]]
*[[Magnetic Force]]
*[[Lorentz Force]]
*[[VPython Modelling of Electric and Magnetic Forces]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Velocity selector====
<div class="mw-collapsible-content">
*[[Lorentz Force]]
*[[Combining Electric and Magnetic Forces]]
</div>
</div>

===Week 10===
<div class="toccolours mw-collapsible mw-collapsed">
====The Hall effect====
<div class="mw-collapsible-content">
*[[Hall Effect]]
*[[Right-Hand Rule]]
*[[Motional Emf]]
*[[Magnetic Force]]
*[[Magnetic Torque]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Motional EMF====
<div class="mw-collapsible-content">
*[[Motional Emf]]
*[[Motional Emf using Faraday's Law]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Magnetic force====
<div class="mw-collapsible-content">
*[[Magnetic Force]]
*[[Lorentz Force]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Magnetic torque====
<div class="mw-collapsible-content">
*[[Magnetic Torque]]
*[[Right-Hand Rule]]
</div>
</div>

===Week 12===

<div class="toccolours mw-collapsible mw-collapsed">
====Gauss's Law====
<div class="mw-collapsible-content">
*[[Gauss's Flux Theorem]]
*[[Gauss's Law]]
*[[Magnetic Flux]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Ampere's Law====
<div class="mw-collapsible-content">
*[[Ampere's Law]]
*[[Ampere-Maxwell Law]]
*[[Magnetic Field of Coaxial Cable Using Ampere's Law]]
*[[Magnetic Field of a Long Thick Wire Using Ampere's Law]]
*[[Magnetic Field of a Toroid Using Ampere's Law]]
*[[Magnetic Field of a Solenoid Using Ampere's Law]]
*[[The Differential Form of Ampere's Law]]
</div>
</div>

===Week 13===
<div class="toccolours mw-collapsible mw-collapsed">
====Semiconductors====
<div class="mw-collapsible-content">
*[[Semiconductor Devices]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Faraday's Law====
<div class="mw-collapsible-content">
*[[Faraday's Law]]
*[[Motional Emf using Faraday's Law]]
*[[Lenz's Law]]

</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">

====Maxwell's equations====
<div class="mw-collapsible-content">
*[[Gauss's Law]]
*[[Magnetic Flux]]
*[[Ampere's Law]]
*[[Faraday's Law]]
*[[Maxwell's Electromagnetic Theory]]
</div>
</div>

===Week 14===
<div class="toccolours mw-collapsible mw-collapsed">
====Circuits revisited====
<div class="mw-collapsible-content">

*[[RLC Circuits]]
*[[LR Circuits]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">

====Inductors====
<div class="mw-collapsible-content">
*[[Inductors]]
*[[Current in an LC Circuit]]
*[[Current in an RL Circuit]]
</div>
</div>

===Week 15===
<div class="toccolours mw-collapsible mw-collapsed">
==== Electromagnetic Radiation ====
<div class="mw-collapsible-content">
*[[Electromagnetic Radiation]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Sparks in the air====
<div class="mw-collapsible-content">
*[[Sparks in Air]]
*[[Spark Plugs]]
</div>
</div>

<div class="toccolours mw-collapsible mw-collapsed">
====Superconductors====
<div class="mw-collapsible-content">
*[[Superconducters]]
*[[Superconductors]]
*[[Meissner effect]]
</div>
</div>
</div>

<div style="float:left; width:30%; padding:1%;">

==Physics 3==

===Week 1===
<div class="toccolours mw-collapsible mw-collapsed">
====Classical Physics====
<div class="mw-collapsible-content">
</div>
</div>

===Week 2===
<div class="toccolours mw-collapsible mw-collapsed">
====Special Relativity====
<div class="mw-collapsible-content">
*[[Frame of Reference]]
*[[Einstein's Theory of Special Relativity]]
*[[Time Dilation]]
*[[Einstein's Theory of General Relativity]]
*[[Albert A. Micheleson & Edward W. Morley]]
*[[Magnetic Force in a Moving Reference Frame]]
</div>
</div>

===Week 3===
<div class="toccolours mw-collapsible mw-collapsed">
====Photons====
<div class="mw-collapsible-content">
*[[Spontaneous Photon Emission]]
*[[Light Scattering: Why is the Sky Blue]]
*[[Lasers]]
*[[Electronic Energy Levels and Photons]]
*[[Quantum Properties of Light]]
</div>
</div>

===Week 4===
<div class="toccolours mw-collapsible mw-collapsed">
====Matter Waves====
<div class="mw-collapsible-content">
*[[Wave-Particle Duality]]
</div>
</div>

===Week 5===
<div class="toccolours mw-collapsible mw-collapsed">
====Wave Mechanics====
<div class="mw-collapsible-content">
*[[Standing Waves]]
*[[Wavelength]]
*[[Wavelength and Frequency]]
*[[Mechanical Waves]]
*[[Transverse and Longitudinal Waves]]
</div>
</div>

===Week 6===
<div class="toccolours mw-collapsible mw-collapsed">
====Rutherford-Bohr Model====
<div class="mw-collapsible-content">
*[[Rutherford Experiment and Atomic Collisions]]
*[[Bohr Model]]
*[[Quantized energy levels]]
*[[Energy graphs and the Bohr model]]
</div>
</div>

===Week 7===
<div class="toccolours mw-collapsible mw-collapsed">
====The Hydrogen Atom====
<div class="mw-collapsible-content">
*[[Quantum Theory]]
*[[Atomic Theory]]
</div>
</div>

===Week 8===
<div class="toccolours mw-collapsible mw-collapsed">
====Many-Electron Atoms====
<div class="mw-collapsible-content">
*[[Quantum Theory]]
*[[Atomic Theory]]
*[[Pauli exclusion principle]]
</div>
</div>

===Week 9===
<div class="toccolours mw-collapsible mw-collapsed">
====Molecules====
<div class="mw-collapsible-content">
</div>
</div>

===Week 10===
<div class="toccolours mw-collapsible mw-collapsed">
====Statistical Physics====
<div class="mw-collapsible-content">
</div>
</div>

===Week 11===
<div class="toccolours mw-collapsible mw-collapsed">
====Condensed Matter Physics====
<div class="mw-collapsible-content">
</div>
</div>

===Week 12===
<div class="toccolours mw-collapsible mw-collapsed">
====The Nucleus====
<div class="mw-collapsible-content">
</div>
</div>

===Week 13===
<div class="toccolours mw-collapsible mw-collapsed">
====Nuclear Physics====
<div class="mw-collapsible-content">
*[[Nuclear Fission]]
*[[Nuclear Energy from Fission and Fusion]]
</div>
</div>

===Week 14===
<div class="toccolours mw-collapsible mw-collapsed">
====Particle Physics====
<div class="mw-collapsible-content">
*[[Elementary Particles and Particle Physics Theory]]
*[[String Theory]]
</div>
</div>
</div>

Potential Difference Path Independence, claimed by Aditya Mohile

2016-11-28T01:46:29Z

Zwright8:

Potential Difference Path Independence, claimed by Aditya Mohile

2016-11-28T01:44:38Z

Zwright8:

''Claimed by Aditya Mohile (Spring 2016)''

'''Claimed by Zachary Wright (Fall 2016)'''

Potential difference or Voltage is a scalar quantity that measures the difference of electric potential energy between two points. Potential energy differences depend only on final and initial states and is independent of the path taken to arrive at those states. Likewise, the potential difference between a final and initial state is independent of the path taken to arrive at those states, therefore the potential difference is said to be path independent. The units of measurement for potential difference is joules per coulomb, which is equal to volts.
==The Main Idea==

The potential difference between two points does not depend on the path taken to get from one point to another because the difference between the locations of the two points remains the same regardless of the path taken to get from one point to another.

===A Mathematical Model===
:<math>\Delta V_{AB} = \int_{A}^{B} \vec{E} \cdot d\vec{l} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>

The potential difference, <math>\Delta V_{AB}</math>, is the integral of the electrical field from point A to point B. This integral, however, is the dot product of <math>\vec{E}</math> and <math>d\vec{l}</math> which is equal to the sum of the negative electric field multiplied by the difference of the x,y and z components of the two points, as see from the equation above. This proves that the path taken from point A to point B does not matter, only the difference in the positions of points A and B are important in calculating the potential difference.

[[File:Electrostatic_definition_of_voltage.jpg]]

As shown above, the potential difference between points A and B in the electric field E is independent of the path.

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===
[[File:Wiki_simple.jpg|center]]
'''Problem''': Locations A, B, and C are in a region of uniform electric field, as shown in the diagram above. Location A is at < -0.6, 0, 0> m. Location B is at < 0.4, 0, 0>. Location C is at < 0.4, -0.3, 0>m. In the region the electric field = < -750, 400, 0> N/C. Calculate the potential difference between A and C traveling from A to B then to C. Calculate the potential difference traveling from A to C directly.

'''Step 1'''
:Calculate the displacement vector from A to B.
:<math>\Delta l = < 0.4, 0, 0> - < -0.6, 0, 0> = < 1, 0, 0></math>
'''Step 2'''
:Calculate the potential difference between A and B.
:<math>\Delta V_{AB} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AB} = -(-750*1 + 400*0 + 0*0) = 750V</math>
'''Step 3'''
:Calculate the displacement vector from B to C.
:<math>\Delta l = < 0.4, -0.3, 0> - < 0.4, 0, 0> = < 0, -0.3, 0></math>
'''Step 4'''
:Calculate the potential difference between B and C.
:<math>\Delta V_{BC} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{BC} = -(-750*0 + 400*-0.3 + 0*0) = 120V</math>
'''Step 5'''
:Add <math>\Delta V_{AB}</math> to <math>\Delta V_{BC}</math> to get <math>\Delta V_{AC}</math>
:<math>\Delta V_{AC} = \Delta V_{AB} + \Delta V_{BC} = 750 + 120 = 870V</math>
Now you have to calculate the potential difference from A to C directly.

'''Step 1'''
:Calculate the displacement vector from A to C.
:<math>\Delta l = < 0.4, -0.3, 0> - < -0.6, 0, 0> = < 1, -0.3, 0></math>
'''Step 2'''
:Calculate the potential difference between A and C.
:<math>\Delta V_{AB} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AB} = -(-750*1 + 400*-0.3 + 0*0) = 750 + 120 = 870V</math>
As you can see, potential difference from A to C directly is the same as the potential difference from A to B then to C.

===Middling===
[[File:Wiki_middling.jpg|350px|center]]
'''Problem''': A capacitor consists of two charged disks of radius 4.7 m separated by a distance s = 2 mm (see the figure). The magnitude of the charge on each disk is 56 µC. Consider points A, B, C, and D inside the capacitor, as shown in the diagram. The distance s1 = 1.5 mm, and the distance s2 = 0.6 mm. (Assume the +x axis is to the right, the +y axis is up, and the +z axis is out.) Find the potential difference between B and D, traveling through A. Find the potential difference from B to D directly.
'''Step 1'''
:Calculate the electric field inside the capacitor.
:<math>\vec E_{cap} = \frac{Q/A}{\epsilon_{0}}</math>
:<math>\vec E_{cap} = \frac{(56*10^{-6})/(\pi*4.7^{2})}{8.85*10^{-12}} = < 91179.912, 0, 0></math>
The electric field only has an x-component because the field travels from the positively charged plate to the negatively charged plate, traveling in the +x direction.

'''Step 2'''
:Calculate the displacement vector from B to A.
:Set the origin at B.
:<math>\Delta l = < s_{1}, 0, 0> - < 0, 0, 0> = < s_{1}, 0, 0> = < 0.0015, 0, 0></math>
'''Step 3'''
:Calculate the potential difference between B and A.
:<math>\Delta V_{BA} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{BA} = -(91179.912*0.0015 + 0*0 + 0*0) = -136.770V</math>
'''Step 4'''
:Calculate the displacement vector from A to D
:<math>\Delta l = < s_{1}, s_{2}, 0> - < s_{1}, 0, 0> = < 0, s_{2}, 0> = < 0, s_{2}, 0> = < 0, 0.006, 0></math>
'''Step 5'''
:Calculate the potential difference between A and D.
:<math>\Delta V_{AD} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AD} = -(91179.912*0 + 0*0 + 0*0) = 0V</math>
'''Step 6'''
:Add <math>\Delta V_{BA}</math> to <math>\Delta V_{AD}</math> to get <math>\Delta V_{BD}</math>
:<math>\Delta V_{BD} = \Delta V_{BA} + \Delta V_{AD} = -136.770 + 0 = -136.770V</math>
Notice that the voltage is negative since we are going from the positively charged plate (high potential) to the negatively charged plate (low potential) losing potential energy in the process.

Now you have to calculate the potential difference from A to C directly.

'''Step 1'''
:Calculate the displacement vector from B to D.
:<math>\Delta l = < s_{1}, s_{2}, 0> - < 0, 0, 0> = < s_{1}, s_{2}, 0> = < s_{1}, s_{2}, 0> = < 0.0015, 0.006, 0></math>
'''Step 2'''
:Calculate the potential difference between B and D.
:<math>\Delta V_{AD} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AD} = -(91179.912*0.0015 + 0*0 + 0*0) = -136.770V</math>
Notice that potential difference traveling from B to A then D is the same as the potential difference traveling from B to D directly. This exemplifies that the potential difference does not depend on the path taken to get from one point to another.

===Difficult===
[[File:Wiki_difficult.png|center]]
'''Problem''': A uniform spherical shell of charge +Q is centered at point B, as shown in the figure below. Show that ΔV = VC − VA is independent of path by calculating ΔV for each of these two paths (actually do the integrals). (Use the following as necessary: Q, R for the radius of the spherical shell, r1 for the radius of the circular arc A → D → C with B at its center, and ε0.)

'''Step 1'''
:Calculate the potential difference from A to C following the path A → B → C.
:First, calculate the potential difference from A to B.
:<math>\Delta V_{AB} = \int_{A}^{B} \vec{E} \cdot d\vec{l}</math>
:In order to calculate the potential difference from A to B, calculate the electric field of the sphere.
:<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}}</math>
:Now plug the electric field calculated above into the integral and solve.
:<math>\Delta V_{AB} = \int_{A}^{B} \vec{E} \cdot d\vec{r} = \int_{r_{1}}^{R} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{R^{2}} - \frac{Q}{r_{1}^{2}})</math>
:Now calculate the potential difference from B to C.
:Use the electric field of the sphere calculated above to solve the below integral.
:<math>\Delta V_{BC} = \int_{B}^{C} \vec{E} \cdot d\vec{l}</math>
:<math>\Delta V_{BC} = \int_{R}^{r_{1}} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{R^{2}})</math>
:Now add <math>\Delta V_{AB}</math> to <math>\Delta V_{BC}</math> to get <math>\Delta V_{AC}</math>
:<math>\Delta V_{AC} = \Delta V_{AB} + \Delta V_{BC} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{R^{2}} - \frac{Q}{r_{1}^{2}}) + \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{R^{2}}) = 0V </math>
'''Step 2'''
:Calculate the potential difference from A to C following the path A → D → C.
:First calculate the potential difference from A to D.
:<math>\Delta V_{AD} = \int_{A}^{D} \vec{E} \cdot d\vec{l}</math>
:Use the same electric field of the sphere calculated in step 1, shown below.
:<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}}</math>
:Now plug the electric field calculated above into the integral and solve.
:<math>\Delta V_{AD} = \int_{A}^{D} \vec{E} \cdot d\vec{r} = \int_{r_{1}}^{r_{1}} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}}) = 0V</math>
:Now calculate the potential difference from D to C.
:Use the electric field of the sphere calculated above to solve the below integral.
:<math>\Delta V_{DC} = \int_{D}^{C} \vec{E} \cdot d\vec{l}</math>
:<math>\Delta V_{DC} = \int_{r_{1}}^{r_{1}} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}})</math>
:Now add <math>\Delta V_{AD}</math> to <math>\Delta V_{DC}</math> to get <math>\Delta V_{AC}</math>
:<math>\Delta V_{AC} = \Delta V_{AD} + \Delta V_{DC} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}}) + \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}}) = 0V </math>
:Notice that either path taken from A to C results in the same potential difference of 0V.

==Connectedness==
#How is this topic connected to something that you are interested in?
::I am very interested in electronics and circuitry so knowing that potential difference is path independent definitely helps in determining the voltage between 2 points in a complicated circuit with many paths.
#How is it connected to your major?
::I am an electrical engineering major so this knowledge is very applicable to determining the voltage between 2 points in a circuit that contains many loops or is very complex.
#In Circuits
::Path Independence of potential difference is critical in solving circuit equations. Many values in a circuit can be calculated using these principles and the Kirchoff's Voltage Law.

==History==

Path independence of potential difference stems from the Law of Conservation of Energy. Many noteworthy scientists worked to affirm this law such as: Isaac Newton, Johann and Daniel Bernoulli, Gottfried Leibniz, and many others.
The unit of measurement for potential difference is joules per coulomb, which is equal to a volt. The unit Volt is named in honor of the Italian scientist Alessandro Volta, who was an innovator in the fields of electricity and power.
Luigi Galvani, an Italian physicist, discovered something he called "animal electricity" which was two different metals connected in series with a frog leg and each other. However, Alessandro Volta, another Italian physicist, realized that the frog leg was acting as a conductor and a detector of electricity, or an electrolyte. He later replaced the frog leg with brine soaked paper and noticed the same results. Through this, Volta discovered electromotive force relating to a galvanic cell, realizing that the difference between the two electrode potentials caused electricity to flow.

== See also ==

https://en.wikipedia.org/wiki/Voltage

===Further reading===

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elewor.html#c2

===External links===

http://mathwiki.ucdavis.edu/Core/Calculus/Vector_Calculus/Integration_in_Vector_Fields/Path_Independence,_Conservative_Fields,_and_Potential_Functions

==References==

Matter & Interactions volume II by Ruth W. Chabay and Bruce A. Sherwood.

[[Category:Energy]]

Potential Difference Path Independence, claimed by Aditya Mohile

2016-11-28T01:36:31Z

Zwright8:

''Claimed by Aditya Mohile (Spring 2016)''

'''Claimed by Zachary Wright (Fall 2016)'''

Potential difference or Voltage is a scalar quantity that measures the difference of electric potential energy between two points. Potential energy differences depend only on final and initial states and is independent of the path taken to arrive at those states. Likewise, the potential difference between a final and initial state is independent of the path taken to arrive at those states, therefore the potential difference is said to be path independent. The units of measurement for potential difference is joules per coulomb, which is equal to volts.
==The Main Idea==

The potential difference between two points does not depend on the path taken to get from one point to another because the difference between the locations of the two points remains the same regardless of the path taken to get from one point to another.

===A Mathematical Model===
:<math>\Delta V_{AB} = \int_{A}^{B} \vec{E} \cdot d\vec{l} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>

The potential difference, <math>\Delta V_{AB}</math>, is the integral of the electrical field from point A to point B. This integral, however, is the dot product of <math>\vec{E}</math> and <math>d\vec{l}</math> which is equal to the sum of the negative electric field multiplied by the difference of the x,y and z components of the two points, as see from the equation above. This proves that the path taken from point A to point B does not matter, only the difference in the positions of points A and B are important in calculating the potential difference.

[[File:Electrostatic_definition_of_voltage.jpg]]

As shown above, the potential difference between points A and B in the electric field E is independent of the path.

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===
[[File:Wiki_simple.jpg|center]]
'''Problem''': Locations A, B, and C are in a region of uniform electric field, as shown in the diagram above. Location A is at < -0.6, 0, 0> m. Location B is at < 0.4, 0, 0>. Location C is at < 0.4, -0.3, 0>m. In the region the electric field = < -750, 400, 0> N/C. Calculate the potential difference between A and C traveling from A to B then to C. Calculate the potential difference traveling from A to C directly.

'''Step 1'''
:Calculate the displacement vector from A to B.
:<math>\Delta l = < 0.4, 0, 0> - < -0.6, 0, 0> = < 1, 0, 0></math>
'''Step 2'''
:Calculate the potential difference between A and B.
:<math>\Delta V_{AB} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AB} = -(-750*1 + 400*0 + 0*0) = 750V</math>
'''Step 3'''
:Calculate the displacement vector from B to C.
:<math>\Delta l = < 0.4, -0.3, 0> - < 0.4, 0, 0> = < 0, -0.3, 0></math>
'''Step 4'''
:Calculate the potential difference between B and C.
:<math>\Delta V_{BC} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{BC} = -(-750*0 + 400*-0.3 + 0*0) = 120V</math>
'''Step 5'''
:Add <math>\Delta V_{AB}</math> to <math>\Delta V_{BC}</math> to get <math>\Delta V_{AC}</math>
:<math>\Delta V_{AC} = \Delta V_{AB} + \Delta V_{BC} = 750 + 120 = 870V</math>
Now you have to calculate the potential difference from A to C directly.

'''Step 1'''
:Calculate the displacement vector from A to C.
:<math>\Delta l = < 0.4, -0.3, 0> - < -0.6, 0, 0> = < 1, -0.3, 0></math>
'''Step 2'''
:Calculate the potential difference between A and C.
:<math>\Delta V_{AB} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AB} = -(-750*1 + 400*-0.3 + 0*0) = 750 + 120 = 870V</math>
As you can see, potential difference from A to C directly is the same as the potential difference from A to B then to C.

===Middling===
[[File:Wiki_middling.jpg|350px|center]]
'''Problem''': A capacitor consists of two charged disks of radius 4.7 m separated by a distance s = 2 mm (see the figure). The magnitude of the charge on each disk is 56 µC. Consider points A, B, C, and D inside the capacitor, as shown in the diagram. The distance s1 = 1.5 mm, and the distance s2 = 0.6 mm. (Assume the +x axis is to the right, the +y axis is up, and the +z axis is out.) Find the potential difference between B and D, traveling through A. Find the potential difference from B to D directly.
'''Step 1'''
:Calculate the electric field inside the capacitor.
:<math>\vec E_{cap} = \frac{Q/A}{\epsilon_{0}}</math>
:<math>\vec E_{cap} = \frac{(56*10^{-6})/(\pi*4.7^{2})}{8.85*10^{-12}} = < 91179.912, 0, 0></math>
The electric field only has an x-component because the field travels from the positively charged plate to the negatively charged plate, traveling in the +x direction.

'''Step 2'''
:Calculate the displacement vector from B to A.
:Set the origin at B.
:<math>\Delta l = < s_{1}, 0, 0> - < 0, 0, 0> = < s_{1}, 0, 0> = < 0.0015, 0, 0></math>
'''Step 3'''
:Calculate the potential difference between B and A.
:<math>\Delta V_{BA} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{BA} = -(91179.912*0.0015 + 0*0 + 0*0) = -136.770V</math>
'''Step 4'''
:Calculate the displacement vector from A to D
:<math>\Delta l = < s_{1}, s_{2}, 0> - < s_{1}, 0, 0> = < 0, s_{2}, 0> = < 0, s_{2}, 0> = < 0, 0.006, 0></math>
'''Step 5'''
:Calculate the potential difference between A and D.
:<math>\Delta V_{AD} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AD} = -(91179.912*0 + 0*0 + 0*0) = 0V</math>
'''Step 6'''
:Add <math>\Delta V_{BA}</math> to <math>\Delta V_{AD}</math> to get <math>\Delta V_{BD}</math>
:<math>\Delta V_{BD} = \Delta V_{BA} + \Delta V_{AD} = -136.770 + 0 = -136.770V</math>
Notice that the voltage is negative since we are going from the positively charged plate (high potential) to the negatively charged plate (low potential) losing potential energy in the process.

Now you have to calculate the potential difference from A to C directly.

'''Step 1'''
:Calculate the displacement vector from B to D.
:<math>\Delta l = < s_{1}, s_{2}, 0> - < 0, 0, 0> = < s_{1}, s_{2}, 0> = < s_{1}, s_{2}, 0> = < 0.0015, 0.006, 0></math>
'''Step 2'''
:Calculate the potential difference between B and D.
:<math>\Delta V_{AD} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AD} = -(91179.912*0.0015 + 0*0 + 0*0) = -136.770V</math>
Notice that potential difference traveling from B to A then D is the same as the potential difference traveling from B to D directly. This exemplifies that the potential difference does not depend on the path taken to get from one point to another.

===Difficult===
[[File:Wiki_difficult.png|center]]
'''Problem''': A uniform spherical shell of charge +Q is centered at point B, as shown in the figure below. Show that ΔV = VC − VA is independent of path by calculating ΔV for each of these two paths (actually do the integrals). (Use the following as necessary: Q, R for the radius of the spherical shell, r1 for the radius of the circular arc A → D → C with B at its center, and ε0.)

'''Step 1'''
:Calculate the potential difference from A to C following the path A → B → C.
:First, calculate the potential difference from A to B.
:<math>\Delta V_{AB} = \int_{A}^{B} \vec{E} \cdot d\vec{l}</math>
:In order to calculate the potential difference from A to B, calculate the electric field of the sphere.
:<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}}</math>
:Now plug the electric field calculated above into the integral and solve.
:<math>\Delta V_{AB} = \int_{A}^{B} \vec{E} \cdot d\vec{r} = \int_{r_{1}}^{R} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{R^{2}} - \frac{Q}{r_{1}^{2}})</math>
:Now calculate the potential difference from B to C.
:Use the electric field of the sphere calculated above to solve the below integral.
:<math>\Delta V_{BC} = \int_{B}^{C} \vec{E} \cdot d\vec{l}</math>
:<math>\Delta V_{BC} = \int_{R}^{r_{1}} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{R^{2}})</math>
:Now add <math>\Delta V_{AB}</math> to <math>\Delta V_{BC}</math> to get <math>\Delta V_{AC}</math>
:<math>\Delta V_{AC} = \Delta V_{AB} + \Delta V_{BC} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{R^{2}} - \frac{Q}{r_{1}^{2}}) + \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{R^{2}}) = 0V </math>
'''Step 2'''
:Calculate the potential difference from A to C following the path A → D → C.
:First calculate the potential difference from A to D.
:<math>\Delta V_{AD} = \int_{A}^{D} \vec{E} \cdot d\vec{l}</math>
:Use the same electric field of the sphere calculated in step 1, shown below.
:<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}}</math>
:Now plug the electric field calculated above into the integral and solve.
:<math>\Delta V_{AD} = \int_{A}^{D} \vec{E} \cdot d\vec{r} = \int_{r_{1}}^{r_{1}} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}}) = 0V</math>
:Now calculate the potential difference from D to C.
:Use the electric field of the sphere calculated above to solve the below integral.
:<math>\Delta V_{DC} = \int_{D}^{C} \vec{E} \cdot d\vec{l}</math>
:<math>\Delta V_{DC} = \int_{r_{1}}^{r_{1}} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}})</math>
:Now add <math>\Delta V_{AD}</math> to <math>\Delta V_{DC}</math> to get <math>\Delta V_{AC}</math>
:<math>\Delta V_{AC} = \Delta V_{AD} + \Delta V_{DC} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}}) + \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}}) = 0V </math>
:Notice that either path taken from A to C results in the same potential difference of 0V.

==Connectedness==
#How is this topic connected to something that you are interested in?
::I am very interested in electronics and circuitry so knowing that potential difference is path independent definitely helps in determining the voltage between 2 points in a complicated circuit with many paths.
#How is it connected to your major?
::I am an electrical engineering major so this knowledge is very applicable to determining the voltage between 2 points in a circuit that contains many loops or is very complex.
#In Circuits
::Path Independence of potential difference is critical in solving circuit equations. Many values in a circuit can be calculated using these principles and the Kirchoff's Voltage Law.

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.
Luigi Galvani, an Italian physicist, discovered something he called "animal electricity" which was two different metals connected in series with a frog leg and each other. However, Alessandro Volta, another Italian physicist, realized that the frog leg was acting as a conductor and a detector of electricity, or an electrolyte. He later replaced the frog leg with brine soaked paper and noticed the same results. Through this, Volta discovered electromotive force relating to a galvanic cell, realizing that the difference between the two electrode potentials caused electricity to flow.

== See also ==

https://en.wikipedia.org/wiki/Voltage

===Further reading===

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elewor.html#c2

===External links===

http://mathwiki.ucdavis.edu/Core/Calculus/Vector_Calculus/Integration_in_Vector_Fields/Path_Independence,_Conservative_Fields,_and_Potential_Functions

==References==

Matter & Interactions volume II by Ruth W. Chabay and Bruce A. Sherwood.

[[Category:Energy]]

Potential Difference Path Independence, claimed by Aditya Mohile

2016-11-28T01:34:06Z

Zwright8:

''Claimed by Aditya Mohile''

'''Claimed by Zachary Wright (Fall 2016)'''

Potential difference or Voltage is a scalar quantity that measures the difference of electric potential energy between two points. Potential energy differences depend only on final and initial states and is independent of the path taken to arrive at those states. Likewise, the potential difference between a final and initial state is independent of the path taken to arrive at those states, therefore the potential difference is said to be path independent. The units of measurement for potential difference is joules per coulomb, which is equal to volts.
==The Main Idea==

The potential difference between two points does not depend on the path taken to get from one point to another because the difference between the locations of the two points remains the same regardless of the path taken to get from one point to another.

===A Mathematical Model===
:<math>\Delta V_{AB} = \int_{A}^{B} \vec{E} \cdot d\vec{l} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>

The potential difference, <math>\Delta V_{AB}</math>, is the integral of the electrical field from point A to point B. This integral, however, is the dot product of <math>\vec{E}</math> and <math>d\vec{l}</math> which is equal to the sum of the negative electric field multiplied by the difference of the x,y and z components of the two points, as see from the equation above. This proves that the path taken from point A to point B does not matter, only the difference in the positions of points A and B are important in calculating the potential difference.

[[File:Electrostatic_definition_of_voltage.jpg]]

As shown above, the potential difference between points A and B in the electric field E is independent of the path.

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===
[[File:Wiki_simple.jpg|center]]
'''Problem''': Locations A, B, and C are in a region of uniform electric field, as shown in the diagram above. Location A is at < -0.6, 0, 0> m. Location B is at < 0.4, 0, 0>. Location C is at < 0.4, -0.3, 0>m. In the region the electric field = < -750, 400, 0> N/C. Calculate the potential difference between A and C traveling from A to B then to C. Calculate the potential difference traveling from A to C directly.

'''Step 1'''
:Calculate the displacement vector from A to B.
:<math>\Delta l = < 0.4, 0, 0> - < -0.6, 0, 0> = < 1, 0, 0></math>
'''Step 2'''
:Calculate the potential difference between A and B.
:<math>\Delta V_{AB} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AB} = -(-750*1 + 400*0 + 0*0) = 750V</math>
'''Step 3'''
:Calculate the displacement vector from B to C.
:<math>\Delta l = < 0.4, -0.3, 0> - < 0.4, 0, 0> = < 0, -0.3, 0></math>
'''Step 4'''
:Calculate the potential difference between B and C.
:<math>\Delta V_{BC} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{BC} = -(-750*0 + 400*-0.3 + 0*0) = 120V</math>
'''Step 5'''
:Add <math>\Delta V_{AB}</math> to <math>\Delta V_{BC}</math> to get <math>\Delta V_{AC}</math>
:<math>\Delta V_{AC} = \Delta V_{AB} + \Delta V_{BC} = 750 + 120 = 870V</math>
Now you have to calculate the potential difference from A to C directly.

'''Step 1'''
:Calculate the displacement vector from A to C.
:<math>\Delta l = < 0.4, -0.3, 0> - < -0.6, 0, 0> = < 1, -0.3, 0></math>
'''Step 2'''
:Calculate the potential difference between A and C.
:<math>\Delta V_{AB} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AB} = -(-750*1 + 400*-0.3 + 0*0) = 750 + 120 = 870V</math>
As you can see, potential difference from A to C directly is the same as the potential difference from A to B then to C.

===Middling===
[[File:Wiki_middling.jpg|350px|center]]
'''Problem''': A capacitor consists of two charged disks of radius 4.7 m separated by a distance s = 2 mm (see the figure). The magnitude of the charge on each disk is 56 µC. Consider points A, B, C, and D inside the capacitor, as shown in the diagram. The distance s1 = 1.5 mm, and the distance s2 = 0.6 mm. (Assume the +x axis is to the right, the +y axis is up, and the +z axis is out.) Find the potential difference between B and D, traveling through A. Find the potential difference from B to D directly.
'''Step 1'''
:Calculate the electric field inside the capacitor.
:<math>\vec E_{cap} = \frac{Q/A}{\epsilon_{0}}</math>
:<math>\vec E_{cap} = \frac{(56*10^{-6})/(\pi*4.7^{2})}{8.85*10^{-12}} = < 91179.912, 0, 0></math>
The electric field only has an x-component because the field travels from the positively charged plate to the negatively charged plate, traveling in the +x direction.

'''Step 2'''
:Calculate the displacement vector from B to A.
:Set the origin at B.
:<math>\Delta l = < s_{1}, 0, 0> - < 0, 0, 0> = < s_{1}, 0, 0> = < 0.0015, 0, 0></math>
'''Step 3'''
:Calculate the potential difference between B and A.
:<math>\Delta V_{BA} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{BA} = -(91179.912*0.0015 + 0*0 + 0*0) = -136.770V</math>
'''Step 4'''
:Calculate the displacement vector from A to D
:<math>\Delta l = < s_{1}, s_{2}, 0> - < s_{1}, 0, 0> = < 0, s_{2}, 0> = < 0, s_{2}, 0> = < 0, 0.006, 0></math>
'''Step 5'''
:Calculate the potential difference between A and D.
:<math>\Delta V_{AD} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AD} = -(91179.912*0 + 0*0 + 0*0) = 0V</math>
'''Step 6'''
:Add <math>\Delta V_{BA}</math> to <math>\Delta V_{AD}</math> to get <math>\Delta V_{BD}</math>
:<math>\Delta V_{BD} = \Delta V_{BA} + \Delta V_{AD} = -136.770 + 0 = -136.770V</math>
Notice that the voltage is negative since we are going from the positively charged plate (high potential) to the negatively charged plate (low potential) losing potential energy in the process.

Now you have to calculate the potential difference from A to C directly.

'''Step 1'''
:Calculate the displacement vector from B to D.
:<math>\Delta l = < s_{1}, s_{2}, 0> - < 0, 0, 0> = < s_{1}, s_{2}, 0> = < s_{1}, s_{2}, 0> = < 0.0015, 0.006, 0></math>
'''Step 2'''
:Calculate the potential difference between B and D.
:<math>\Delta V_{AD} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AD} = -(91179.912*0.0015 + 0*0 + 0*0) = -136.770V</math>
Notice that potential difference traveling from B to A then D is the same as the potential difference traveling from B to D directly. This exemplifies that the potential difference does not depend on the path taken to get from one point to another.

===Difficult===
[[File:Wiki_difficult.png|center]]
'''Problem''': A uniform spherical shell of charge +Q is centered at point B, as shown in the figure below. Show that ΔV = VC − VA is independent of path by calculating ΔV for each of these two paths (actually do the integrals). (Use the following as necessary: Q, R for the radius of the spherical shell, r1 for the radius of the circular arc A → D → C with B at its center, and ε0.)

'''Step 1'''
:Calculate the potential difference from A to C following the path A → B → C.
:First, calculate the potential difference from A to B.
:<math>\Delta V_{AB} = \int_{A}^{B} \vec{E} \cdot d\vec{l}</math>
:In order to calculate the potential difference from A to B, calculate the electric field of the sphere.
:<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}}</math>
:Now plug the electric field calculated above into the integral and solve.
:<math>\Delta V_{AB} = \int_{A}^{B} \vec{E} \cdot d\vec{r} = \int_{r_{1}}^{R} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{R^{2}} - \frac{Q}{r_{1}^{2}})</math>
:Now calculate the potential difference from B to C.
:Use the electric field of the sphere calculated above to solve the below integral.
:<math>\Delta V_{BC} = \int_{B}^{C} \vec{E} \cdot d\vec{l}</math>
:<math>\Delta V_{BC} = \int_{R}^{r_{1}} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{R^{2}})</math>
:Now add <math>\Delta V_{AB}</math> to <math>\Delta V_{BC}</math> to get <math>\Delta V_{AC}</math>
:<math>\Delta V_{AC} = \Delta V_{AB} + \Delta V_{BC} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{R^{2}} - \frac{Q}{r_{1}^{2}}) + \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{R^{2}}) = 0V </math>
'''Step 2'''
:Calculate the potential difference from A to C following the path A → D → C.
:First calculate the potential difference from A to D.
:<math>\Delta V_{AD} = \int_{A}^{D} \vec{E} \cdot d\vec{l}</math>
:Use the same electric field of the sphere calculated in step 1, shown below.
:<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}}</math>
:Now plug the electric field calculated above into the integral and solve.
:<math>\Delta V_{AD} = \int_{A}^{D} \vec{E} \cdot d\vec{r} = \int_{r_{1}}^{r_{1}} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}}) = 0V</math>
:Now calculate the potential difference from D to C.
:Use the electric field of the sphere calculated above to solve the below integral.
:<math>\Delta V_{DC} = \int_{D}^{C} \vec{E} \cdot d\vec{l}</math>
:<math>\Delta V_{DC} = \int_{r_{1}}^{r_{1}} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}})</math>
:Now add <math>\Delta V_{AD}</math> to <math>\Delta V_{DC}</math> to get <math>\Delta V_{AC}</math>
:<math>\Delta V_{AC} = \Delta V_{AD} + \Delta V_{DC} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}}) + \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}}) = 0V </math>
:Notice that either path taken from A to C results in the same potential difference of 0V.

==Connectedness==
#How is this topic connected to something that you are interested in?
::I am very interested in electronics and circuitry so knowing that potential difference is path independent definitely helps in determining the voltage between 2 points in a complicated circuit with many paths.
#How is it connected to your major?
::I am an electrical engineering major so this knowledge is very applicable to determining the voltage between 2 points in a circuit that contains many loops or is very complex.
#In Circuits
::Path Independence of potential difference is critical in solving circuit equations. Many values in a circuit can be calculated using these principles and the Kirchoff's Voltage Law.

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.
Luigi Galvani, an Italian physicist, discovered something he called "animal electricity" which was two different metals connected in series with a frog leg and each other. However, Alessandro Volta, another Italian physicist, realized that the frog leg was acting as a conductor and a detector of electricity, or an electrolyte. He later replaced the frog leg with brine soaked paper and noticed the same results. Through this, Volta discovered electromotive force relating to a galvanic cell, realizing that the difference between the two electrode potentials caused electricity to flow.

== See also ==

https://en.wikipedia.org/wiki/Voltage

===Further reading===

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elewor.html#c2

===External links===

http://mathwiki.ucdavis.edu/Core/Calculus/Vector_Calculus/Integration_in_Vector_Fields/Path_Independence,_Conservative_Fields,_and_Potential_Functions

==References==

Matter & Interactions volume II by Ruth W. Chabay and Bruce A. Sherwood.

[[Category:Energy]]

Potential Difference Path Independence, claimed by Aditya Mohile

2016-11-28T01:20:16Z

Zwright8:

''Claimed by Aditya Mohile''

'''Claimed by Zachary Wright (Fall 2016)'''

Potential difference or Voltage is a scalar quantity that measures the difference of electric potential energy between two points. Potential energy differences depend only on final and initial states and is independent of the path taken to arrive at those states. Likewise, the potential difference between a final and initial state is independent of the path taken to arrive at those states, therefore the potential difference is said to be path independent. The units of measurement for potential difference is joules per coulomb, which is equal to volts.
==The Main Idea==

The potential difference between two points does not depend on the path taken to get from one point to another because the difference between the locations of the two points remains the same regardless of the path taken to get from one point to another.

===A Mathematical Model===
:<math>\Delta V_{AB} = \int_{A}^{B} \vec{E} \cdot d\vec{l} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>

The potential difference, <math>\Delta V_{AB}</math>, is the integral of the electrical field from point A to point B. This integral, however, is the dot product of <math>\vec{E}</math> and <math>d\vec{l}</math> which is equal to the sum of the negative electric field multiplied by the difference of the x,y and z components of the two points, as see from the equation above. This proves that the path taken from point A to point B does not matter, only the difference in the positions of points A and B are important in calculating the potential difference.

[[File:Electrostatic_definition_of_voltage.jpg]]

As shown above, the potential difference between points A and B in the electric field E is independent of the path.

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===
[[File:Wiki_simple.jpg|center]]
'''Problem''': Locations A, B, and C are in a region of uniform electric field, as shown in the diagram above. Location A is at < -0.6, 0, 0> m. Location B is at < 0.4, 0, 0>. Location C is at < 0.4, -0.3, 0>m. In the region the electric field = < -750, 400, 0> N/C. Calculate the potential difference between A and C traveling from A to B then to C. Calculate the potential difference traveling from A to C directly.

'''Step 1'''
:Calculate the displacement vector from A to B.
:<math>\Delta l = < 0.4, 0, 0> - < -0.6, 0, 0> = < 1, 0, 0></math>
'''Step 2'''
:Calculate the potential difference between A and B.
:<math>\Delta V_{AB} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AB} = -(-750*1 + 400*0 + 0*0) = 750V</math>
'''Step 3'''
:Calculate the displacement vector from B to C.
:<math>\Delta l = < 0.4, -0.3, 0> - < 0.4, 0, 0> = < 0, -0.3, 0></math>
'''Step 4'''
:Calculate the potential difference between B and C.
:<math>\Delta V_{BC} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{BC} = -(-750*0 + 400*-0.3 + 0*0) = 120V</math>
'''Step 5'''
:Add <math>\Delta V_{AB}</math> to <math>\Delta V_{BC}</math> to get <math>\Delta V_{AC}</math>
:<math>\Delta V_{AC} = \Delta V_{AB} + \Delta V_{BC} = 750 + 120 = 870V</math>
Now you have to calculate the potential difference from A to C directly.

'''Step 1'''
:Calculate the displacement vector from A to C.
:<math>\Delta l = < 0.4, -0.3, 0> - < -0.6, 0, 0> = < 1, -0.3, 0></math>
'''Step 2'''
:Calculate the potential difference between A and C.
:<math>\Delta V_{AB} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AB} = -(-750*1 + 400*-0.3 + 0*0) = 750 + 120 = 870V</math>
As you can see, potential difference from A to C directly is the same as the potential difference from A to B then to C.

===Middling===
[[File:Wiki_middling.jpg|350px|center]]
'''Problem''': A capacitor consists of two charged disks of radius 4.7 m separated by a distance s = 2 mm (see the figure). The magnitude of the charge on each disk is 56 µC. Consider points A, B, C, and D inside the capacitor, as shown in the diagram. The distance s1 = 1.5 mm, and the distance s2 = 0.6 mm. (Assume the +x axis is to the right, the +y axis is up, and the +z axis is out.) Find the potential difference between B and D, traveling through A. Find the potential difference from B to D directly.
'''Step 1'''
:Calculate the electric field inside the capacitor.
:<math>\vec E_{cap} = \frac{Q/A}{\epsilon_{0}}</math>
:<math>\vec E_{cap} = \frac{(56*10^{-6})/(\pi*4.7^{2})}{8.85*10^{-12}} = < 91179.912, 0, 0></math>
The electric field only has an x-component because the field travels from the positively charged plate to the negatively charged plate, traveling in the +x direction.

'''Step 2'''
:Calculate the displacement vector from B to A.
:Set the origin at B.
:<math>\Delta l = < s_{1}, 0, 0> - < 0, 0, 0> = < s_{1}, 0, 0> = < 0.0015, 0, 0></math>
'''Step 3'''
:Calculate the potential difference between B and A.
:<math>\Delta V_{BA} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{BA} = -(91179.912*0.0015 + 0*0 + 0*0) = -136.770V</math>
'''Step 4'''
:Calculate the displacement vector from A to D
:<math>\Delta l = < s_{1}, s_{2}, 0> - < s_{1}, 0, 0> = < 0, s_{2}, 0> = < 0, s_{2}, 0> = < 0, 0.006, 0></math>
'''Step 5'''
:Calculate the potential difference between A and D.
:<math>\Delta V_{AD} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AD} = -(91179.912*0 + 0*0 + 0*0) = 0V</math>
'''Step 6'''
:Add <math>\Delta V_{BA}</math> to <math>\Delta V_{AD}</math> to get <math>\Delta V_{BD}</math>
:<math>\Delta V_{BD} = \Delta V_{BA} + \Delta V_{AD} = -136.770 + 0 = -136.770V</math>
Notice that the voltage is negative since we are going from the positively charged plate (high potential) to the negatively charged plate (low potential) losing potential energy in the process.

Now you have to calculate the potential difference from A to C directly.

'''Step 1'''
:Calculate the displacement vector from B to D.
:<math>\Delta l = < s_{1}, s_{2}, 0> - < 0, 0, 0> = < s_{1}, s_{2}, 0> = < s_{1}, s_{2}, 0> = < 0.0015, 0.006, 0></math>
'''Step 2'''
:Calculate the potential difference between B and D.
:<math>\Delta V_{AD} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AD} = -(91179.912*0.0015 + 0*0 + 0*0) = -136.770V</math>
Notice that potential difference traveling from B to A then D is the same as the potential difference traveling from B to D directly. This exemplifies that the potential difference does not depend on the path taken to get from one point to another.

===Difficult===
[[File:Wiki_difficult.png|center]]
'''Problem''': A uniform spherical shell of charge +Q is centered at point B, as shown in the figure below. Show that ΔV = VC − VA is independent of path by calculating ΔV for each of these two paths (actually do the integrals). (Use the following as necessary: Q, R for the radius of the spherical shell, r1 for the radius of the circular arc A → D → C with B at its center, and ε0.)

'''Step 1'''
:Calculate the potential difference from A to C following the path A → B → C.
:First, calculate the potential difference from A to B.
:<math>\Delta V_{AB} = \int_{A}^{B} \vec{E} \cdot d\vec{l}</math>
:In order to calculate the potential difference from A to B, calculate the electric field of the sphere.
:<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}}</math>
:Now plug the electric field calculated above into the integral and solve.
:<math>\Delta V_{AB} = \int_{A}^{B} \vec{E} \cdot d\vec{r} = \int_{r_{1}}^{R} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{R^{2}} - \frac{Q}{r_{1}^{2}})</math>
:Now calculate the potential difference from B to C.
:Use the electric field of the sphere calculated above to solve the below integral.
:<math>\Delta V_{BC} = \int_{B}^{C} \vec{E} \cdot d\vec{l}</math>
:<math>\Delta V_{BC} = \int_{R}^{r_{1}} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{R^{2}})</math>
:Now add <math>\Delta V_{AB}</math> to <math>\Delta V_{BC}</math> to get <math>\Delta V_{AC}</math>
:<math>\Delta V_{AC} = \Delta V_{AB} + \Delta V_{BC} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{R^{2}} - \frac{Q}{r_{1}^{2}}) + \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{R^{2}}) = 0V </math>
'''Step 2'''
:Calculate the potential difference from A to C following the path A → D → C.
:First calculate the potential difference from A to D.
:<math>\Delta V_{AD} = \int_{A}^{D} \vec{E} \cdot d\vec{l}</math>
:Use the same electric field of the sphere calculated in step 1, shown below.
:<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}}</math>
:Now plug the electric field calculated above into the integral and solve.
:<math>\Delta V_{AD} = \int_{A}^{D} \vec{E} \cdot d\vec{r} = \int_{r_{1}}^{r_{1}} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}}) = 0V</math>
:Now calculate the potential difference from D to C.
:Use the electric field of the sphere calculated above to solve the below integral.
:<math>\Delta V_{DC} = \int_{D}^{C} \vec{E} \cdot d\vec{l}</math>
:<math>\Delta V_{DC} = \int_{r_{1}}^{r_{1}} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}})</math>
:Now add <math>\Delta V_{AD}</math> to <math>\Delta V_{DC}</math> to get <math>\Delta V_{AC}</math>
:<math>\Delta V_{AC} = \Delta V_{AD} + \Delta V_{DC} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}}) + \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}}) = 0V </math>
:Notice that either path taken from A to C results in the same potential difference of 0V.

==Connectedness==
#How is this topic connected to something that you are interested in?
::I am very interested in electronics and circuitry so knowing that potential difference is path independent definitely helps in determining the voltage between 2 points in a complicated circuit with many paths.
#How is it connected to your major?
::I am an electrical engineering major so this knowledge is very applicable to determining the voltage between 2 points in a circuit that contains many loops or is very complex.

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.
Luigi Galvani, an Italian physicist, discovered something he called "animal electricity" which was two different metals connected in series with a frog leg and each other. However, Alessandro Volta, another Italian physicist, realized that the frog leg was acting as a conductor and a detector of electricity, or an electrolyte. He later replaced the frog leg with brine soaked paper and noticed the same results. Through this, Volta discovered electromotive force relating to a galvanic cell, realizing that the difference between the two electrode potentials caused electricity to flow.

== See also ==

https://en.wikipedia.org/wiki/Voltage

===Further reading===

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elewor.html#c2

===External links===

http://mathwiki.ucdavis.edu/Core/Calculus/Vector_Calculus/Integration_in_Vector_Fields/Path_Independence,_Conservative_Fields,_and_Potential_Functions

==References==

Matter & Interactions volume II by Ruth W. Chabay and Bruce A. Sherwood.

[[Category:Energy]]

Potential Difference Path Independence, claimed by Aditya Mohile

2016-11-28T01:19:04Z

Zwright8:

''Claimed by Aditya Mohile''

'''Claimed by Zachary Wright (Fall 2016)'''

Potential difference or Voltage is a scalar quantity that measures the difference of electric potential energy between two points. Potential energy differences depend only on final and initial states and is independent of the path taken to arrive at those states. Likewise, the potential difference between a final and initial state is independent of the path taken to arrive at those states, therefore the potential difference is said to be path independent. The units of measurement for potential difference is joules per coulomb, which is equal to volts.
==The Main Idea==

The potential difference between two points does not depend on the path taken to get from one point to another because the difference between the locations of the two points remains the same regardless of the path taken to get from one point to another.

===A Mathematical Model===
:<math>\Delta V_{AB} = \int_{A}^{B} \vec{E} \cdot d\vec{l} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>

The potential difference, <math>\Delta V_{AB}</math>, is the integral of the electrical field from point A to point B. This integral, however, is the dot product of <math>\vec{E}</math> and <math>d\vec{l}</math> which is equal to the sum of the negative electric field multiplied by the difference of the x,y and z components of the two points, as see from the equation above. This proves that the path taken from point A to point B does not matter, only the difference in the positions of points A and B are important in calculating the potential difference.

[[File:Electrostatic_definition_of_voltage.jpg]]

As shown above, the potential difference is independent of the path.

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===
[[File:Wiki_simple.jpg|center]]
'''Problem''': Locations A, B, and C are in a region of uniform electric field, as shown in the diagram above. Location A is at < -0.6, 0, 0> m. Location B is at < 0.4, 0, 0>. Location C is at < 0.4, -0.3, 0>m. In the region the electric field = < -750, 400, 0> N/C. Calculate the potential difference between A and C traveling from A to B then to C. Calculate the potential difference traveling from A to C directly.

'''Step 1'''
:Calculate the displacement vector from A to B.
:<math>\Delta l = < 0.4, 0, 0> - < -0.6, 0, 0> = < 1, 0, 0></math>
'''Step 2'''
:Calculate the potential difference between A and B.
:<math>\Delta V_{AB} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AB} = -(-750*1 + 400*0 + 0*0) = 750V</math>
'''Step 3'''
:Calculate the displacement vector from B to C.
:<math>\Delta l = < 0.4, -0.3, 0> - < 0.4, 0, 0> = < 0, -0.3, 0></math>
'''Step 4'''
:Calculate the potential difference between B and C.
:<math>\Delta V_{BC} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{BC} = -(-750*0 + 400*-0.3 + 0*0) = 120V</math>
'''Step 5'''
:Add <math>\Delta V_{AB}</math> to <math>\Delta V_{BC}</math> to get <math>\Delta V_{AC}</math>
:<math>\Delta V_{AC} = \Delta V_{AB} + \Delta V_{BC} = 750 + 120 = 870V</math>
Now you have to calculate the potential difference from A to C directly.

'''Step 1'''
:Calculate the displacement vector from A to C.
:<math>\Delta l = < 0.4, -0.3, 0> - < -0.6, 0, 0> = < 1, -0.3, 0></math>
'''Step 2'''
:Calculate the potential difference between A and C.
:<math>\Delta V_{AB} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AB} = -(-750*1 + 400*-0.3 + 0*0) = 750 + 120 = 870V</math>
As you can see, potential difference from A to C directly is the same as the potential difference from A to B then to C.

===Middling===
[[File:Wiki_middling.jpg|350px|center]]
'''Problem''': A capacitor consists of two charged disks of radius 4.7 m separated by a distance s = 2 mm (see the figure). The magnitude of the charge on each disk is 56 µC. Consider points A, B, C, and D inside the capacitor, as shown in the diagram. The distance s1 = 1.5 mm, and the distance s2 = 0.6 mm. (Assume the +x axis is to the right, the +y axis is up, and the +z axis is out.) Find the potential difference between B and D, traveling through A. Find the potential difference from B to D directly.
'''Step 1'''
:Calculate the electric field inside the capacitor.
:<math>\vec E_{cap} = \frac{Q/A}{\epsilon_{0}}</math>
:<math>\vec E_{cap} = \frac{(56*10^{-6})/(\pi*4.7^{2})}{8.85*10^{-12}} = < 91179.912, 0, 0></math>
The electric field only has an x-component because the field travels from the positively charged plate to the negatively charged plate, traveling in the +x direction.

'''Step 2'''
:Calculate the displacement vector from B to A.
:Set the origin at B.
:<math>\Delta l = < s_{1}, 0, 0> - < 0, 0, 0> = < s_{1}, 0, 0> = < 0.0015, 0, 0></math>
'''Step 3'''
:Calculate the potential difference between B and A.
:<math>\Delta V_{BA} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{BA} = -(91179.912*0.0015 + 0*0 + 0*0) = -136.770V</math>
'''Step 4'''
:Calculate the displacement vector from A to D
:<math>\Delta l = < s_{1}, s_{2}, 0> - < s_{1}, 0, 0> = < 0, s_{2}, 0> = < 0, s_{2}, 0> = < 0, 0.006, 0></math>
'''Step 5'''
:Calculate the potential difference between A and D.
:<math>\Delta V_{AD} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AD} = -(91179.912*0 + 0*0 + 0*0) = 0V</math>
'''Step 6'''
:Add <math>\Delta V_{BA}</math> to <math>\Delta V_{AD}</math> to get <math>\Delta V_{BD}</math>
:<math>\Delta V_{BD} = \Delta V_{BA} + \Delta V_{AD} = -136.770 + 0 = -136.770V</math>
Notice that the voltage is negative since we are going from the positively charged plate (high potential) to the negatively charged plate (low potential) losing potential energy in the process.

Now you have to calculate the potential difference from A to C directly.

'''Step 1'''
:Calculate the displacement vector from B to D.
:<math>\Delta l = < s_{1}, s_{2}, 0> - < 0, 0, 0> = < s_{1}, s_{2}, 0> = < s_{1}, s_{2}, 0> = < 0.0015, 0.006, 0></math>
'''Step 2'''
:Calculate the potential difference between B and D.
:<math>\Delta V_{AD} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AD} = -(91179.912*0.0015 + 0*0 + 0*0) = -136.770V</math>
Notice that potential difference traveling from B to A then D is the same as the potential difference traveling from B to D directly. This exemplifies that the potential difference does not depend on the path taken to get from one point to another.

===Difficult===
[[File:Wiki_difficult.png|center]]
'''Problem''': A uniform spherical shell of charge +Q is centered at point B, as shown in the figure below. Show that ΔV = VC − VA is independent of path by calculating ΔV for each of these two paths (actually do the integrals). (Use the following as necessary: Q, R for the radius of the spherical shell, r1 for the radius of the circular arc A → D → C with B at its center, and ε0.)

'''Step 1'''
:Calculate the potential difference from A to C following the path A → B → C.
:First, calculate the potential difference from A to B.
:<math>\Delta V_{AB} = \int_{A}^{B} \vec{E} \cdot d\vec{l}</math>
:In order to calculate the potential difference from A to B, calculate the electric field of the sphere.
:<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}}</math>
:Now plug the electric field calculated above into the integral and solve.
:<math>\Delta V_{AB} = \int_{A}^{B} \vec{E} \cdot d\vec{r} = \int_{r_{1}}^{R} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{R^{2}} - \frac{Q}{r_{1}^{2}})</math>
:Now calculate the potential difference from B to C.
:Use the electric field of the sphere calculated above to solve the below integral.
:<math>\Delta V_{BC} = \int_{B}^{C} \vec{E} \cdot d\vec{l}</math>
:<math>\Delta V_{BC} = \int_{R}^{r_{1}} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{R^{2}})</math>
:Now add <math>\Delta V_{AB}</math> to <math>\Delta V_{BC}</math> to get <math>\Delta V_{AC}</math>
:<math>\Delta V_{AC} = \Delta V_{AB} + \Delta V_{BC} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{R^{2}} - \frac{Q}{r_{1}^{2}}) + \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{R^{2}}) = 0V </math>
'''Step 2'''
:Calculate the potential difference from A to C following the path A → D → C.
:First calculate the potential difference from A to D.
:<math>\Delta V_{AD} = \int_{A}^{D} \vec{E} \cdot d\vec{l}</math>
:Use the same electric field of the sphere calculated in step 1, shown below.
:<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}}</math>
:Now plug the electric field calculated above into the integral and solve.
:<math>\Delta V_{AD} = \int_{A}^{D} \vec{E} \cdot d\vec{r} = \int_{r_{1}}^{r_{1}} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}}) = 0V</math>
:Now calculate the potential difference from D to C.
:Use the electric field of the sphere calculated above to solve the below integral.
:<math>\Delta V_{DC} = \int_{D}^{C} \vec{E} \cdot d\vec{l}</math>
:<math>\Delta V_{DC} = \int_{r_{1}}^{r_{1}} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}})</math>
:Now add <math>\Delta V_{AD}</math> to <math>\Delta V_{DC}</math> to get <math>\Delta V_{AC}</math>
:<math>\Delta V_{AC} = \Delta V_{AD} + \Delta V_{DC} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}}) + \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}}) = 0V </math>
:Notice that either path taken from A to C results in the same potential difference of 0V.

==Connectedness==
#How is this topic connected to something that you are interested in?
::I am very interested in electronics and circuitry so knowing that potential difference is path independent definitely helps in determining the voltage between 2 points in a complicated circuit with many paths.
#How is it connected to your major?
::I am an electrical engineering major so this knowledge is very applicable to determining the voltage between 2 points in a circuit that contains many loops or is very complex.

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.
Luigi Galvani, an Italian physicist, discovered something he called "animal electricity" which was two different metals connected in series with a frog leg and each other. However, Alessandro Volta, another Italian physicist, realized that the frog leg was acting as a conductor and a detector of electricity, or an electrolyte. He later replaced the frog leg with brine soaked paper and noticed the same results. Through this, Volta discovered electromotive force relating to a galvanic cell, realizing that the difference between the two electrode potentials caused electricity to flow.

== See also ==

https://en.wikipedia.org/wiki/Voltage

===Further reading===

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elewor.html#c2

===External links===

http://mathwiki.ucdavis.edu/Core/Calculus/Vector_Calculus/Integration_in_Vector_Fields/Path_Independence,_Conservative_Fields,_and_Potential_Functions

==References==

Matter & Interactions volume II by Ruth W. Chabay and Bruce A. Sherwood.

[[Category:Energy]]

Potential Difference Path Independence, claimed by Aditya Mohile

2016-11-28T01:18:34Z

Zwright8:

''Claimed by Aditya Mohile''

'''Claimed by Zachary Wright (Fall 2016)'''

Potential difference or Voltage is a scalar quantity that measures the difference of electric potential energy between two points. Potential energy differences depend only on final and initial states and is independent of the path taken to arrive at those states. Likewise, the potential difference between a final and initial state is independent of the path taken to arrive at those states, therefore the potential difference is said to be path independent. The units of measurement for potential difference is joules per coulomb, which is equal to volts.
==The Main Idea==

The potential difference between two points does not depend on the path taken to get from one point to another because the difference between the locations of the two points remains the same regardless of the path taken to get from one point to another.

===A Mathematical Model===
:<math>\Delta V_{AB} = \int_{A}^{B} \vec{E} \cdot d\vec{l} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>

The potential difference, <math>\Delta V_{AB}</math>, is the integral of the electrical field from point A to point B. This integral, however, is the dot product of <math>\vec{E}</math> and <math>d\vec{l}</math> which is equal to the sum of the negative electric field multiplied by the difference of the x,y and z components of the two points, as see from the equation above. This proves that the path taken from point A to point B does not matter, only the difference in the positions of points A and B are important in calculating the potential difference.

[[File:Electrostatic_definition_of_voltage.jpg]]
As shown above, the potential difference is independent of the path.

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===
[[File:Wiki_simple.jpg|center]]
'''Problem''': Locations A, B, and C are in a region of uniform electric field, as shown in the diagram above. Location A is at < -0.6, 0, 0> m. Location B is at < 0.4, 0, 0>. Location C is at < 0.4, -0.3, 0>m. In the region the electric field = < -750, 400, 0> N/C. Calculate the potential difference between A and C traveling from A to B then to C. Calculate the potential difference traveling from A to C directly.

'''Step 1'''
:Calculate the displacement vector from A to B.
:<math>\Delta l = < 0.4, 0, 0> - < -0.6, 0, 0> = < 1, 0, 0></math>
'''Step 2'''
:Calculate the potential difference between A and B.
:<math>\Delta V_{AB} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AB} = -(-750*1 + 400*0 + 0*0) = 750V</math>
'''Step 3'''
:Calculate the displacement vector from B to C.
:<math>\Delta l = < 0.4, -0.3, 0> - < 0.4, 0, 0> = < 0, -0.3, 0></math>
'''Step 4'''
:Calculate the potential difference between B and C.
:<math>\Delta V_{BC} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{BC} = -(-750*0 + 400*-0.3 + 0*0) = 120V</math>
'''Step 5'''
:Add <math>\Delta V_{AB}</math> to <math>\Delta V_{BC}</math> to get <math>\Delta V_{AC}</math>
:<math>\Delta V_{AC} = \Delta V_{AB} + \Delta V_{BC} = 750 + 120 = 870V</math>
Now you have to calculate the potential difference from A to C directly.

'''Step 1'''
:Calculate the displacement vector from A to C.
:<math>\Delta l = < 0.4, -0.3, 0> - < -0.6, 0, 0> = < 1, -0.3, 0></math>
'''Step 2'''
:Calculate the potential difference between A and C.
:<math>\Delta V_{AB} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AB} = -(-750*1 + 400*-0.3 + 0*0) = 750 + 120 = 870V</math>
As you can see, potential difference from A to C directly is the same as the potential difference from A to B then to C.

===Middling===
[[File:Wiki_middling.jpg|350px|center]]
'''Problem''': A capacitor consists of two charged disks of radius 4.7 m separated by a distance s = 2 mm (see the figure). The magnitude of the charge on each disk is 56 µC. Consider points A, B, C, and D inside the capacitor, as shown in the diagram. The distance s1 = 1.5 mm, and the distance s2 = 0.6 mm. (Assume the +x axis is to the right, the +y axis is up, and the +z axis is out.) Find the potential difference between B and D, traveling through A. Find the potential difference from B to D directly.
'''Step 1'''
:Calculate the electric field inside the capacitor.
:<math>\vec E_{cap} = \frac{Q/A}{\epsilon_{0}}</math>
:<math>\vec E_{cap} = \frac{(56*10^{-6})/(\pi*4.7^{2})}{8.85*10^{-12}} = < 91179.912, 0, 0></math>
The electric field only has an x-component because the field travels from the positively charged plate to the negatively charged plate, traveling in the +x direction.

'''Step 2'''
:Calculate the displacement vector from B to A.
:Set the origin at B.
:<math>\Delta l = < s_{1}, 0, 0> - < 0, 0, 0> = < s_{1}, 0, 0> = < 0.0015, 0, 0></math>
'''Step 3'''
:Calculate the potential difference between B and A.
:<math>\Delta V_{BA} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{BA} = -(91179.912*0.0015 + 0*0 + 0*0) = -136.770V</math>
'''Step 4'''
:Calculate the displacement vector from A to D
:<math>\Delta l = < s_{1}, s_{2}, 0> - < s_{1}, 0, 0> = < 0, s_{2}, 0> = < 0, s_{2}, 0> = < 0, 0.006, 0></math>
'''Step 5'''
:Calculate the potential difference between A and D.
:<math>\Delta V_{AD} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AD} = -(91179.912*0 + 0*0 + 0*0) = 0V</math>
'''Step 6'''
:Add <math>\Delta V_{BA}</math> to <math>\Delta V_{AD}</math> to get <math>\Delta V_{BD}</math>
:<math>\Delta V_{BD} = \Delta V_{BA} + \Delta V_{AD} = -136.770 + 0 = -136.770V</math>
Notice that the voltage is negative since we are going from the positively charged plate (high potential) to the negatively charged plate (low potential) losing potential energy in the process.

Now you have to calculate the potential difference from A to C directly.

'''Step 1'''
:Calculate the displacement vector from B to D.
:<math>\Delta l = < s_{1}, s_{2}, 0> - < 0, 0, 0> = < s_{1}, s_{2}, 0> = < s_{1}, s_{2}, 0> = < 0.0015, 0.006, 0></math>
'''Step 2'''
:Calculate the potential difference between B and D.
:<math>\Delta V_{AD} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AD} = -(91179.912*0.0015 + 0*0 + 0*0) = -136.770V</math>
Notice that potential difference traveling from B to A then D is the same as the potential difference traveling from B to D directly. This exemplifies that the potential difference does not depend on the path taken to get from one point to another.

===Difficult===
[[File:Wiki_difficult.png|center]]
'''Problem''': A uniform spherical shell of charge +Q is centered at point B, as shown in the figure below. Show that ΔV = VC − VA is independent of path by calculating ΔV for each of these two paths (actually do the integrals). (Use the following as necessary: Q, R for the radius of the spherical shell, r1 for the radius of the circular arc A → D → C with B at its center, and ε0.)

'''Step 1'''
:Calculate the potential difference from A to C following the path A → B → C.
:First, calculate the potential difference from A to B.
:<math>\Delta V_{AB} = \int_{A}^{B} \vec{E} \cdot d\vec{l}</math>
:In order to calculate the potential difference from A to B, calculate the electric field of the sphere.
:<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}}</math>
:Now plug the electric field calculated above into the integral and solve.
:<math>\Delta V_{AB} = \int_{A}^{B} \vec{E} \cdot d\vec{r} = \int_{r_{1}}^{R} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{R^{2}} - \frac{Q}{r_{1}^{2}})</math>
:Now calculate the potential difference from B to C.
:Use the electric field of the sphere calculated above to solve the below integral.
:<math>\Delta V_{BC} = \int_{B}^{C} \vec{E} \cdot d\vec{l}</math>
:<math>\Delta V_{BC} = \int_{R}^{r_{1}} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{R^{2}})</math>
:Now add <math>\Delta V_{AB}</math> to <math>\Delta V_{BC}</math> to get <math>\Delta V_{AC}</math>
:<math>\Delta V_{AC} = \Delta V_{AB} + \Delta V_{BC} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{R^{2}} - \frac{Q}{r_{1}^{2}}) + \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{R^{2}}) = 0V </math>
'''Step 2'''
:Calculate the potential difference from A to C following the path A → D → C.
:First calculate the potential difference from A to D.
:<math>\Delta V_{AD} = \int_{A}^{D} \vec{E} \cdot d\vec{l}</math>
:Use the same electric field of the sphere calculated in step 1, shown below.
:<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}}</math>
:Now plug the electric field calculated above into the integral and solve.
:<math>\Delta V_{AD} = \int_{A}^{D} \vec{E} \cdot d\vec{r} = \int_{r_{1}}^{r_{1}} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}}) = 0V</math>
:Now calculate the potential difference from D to C.
:Use the electric field of the sphere calculated above to solve the below integral.
:<math>\Delta V_{DC} = \int_{D}^{C} \vec{E} \cdot d\vec{l}</math>
:<math>\Delta V_{DC} = \int_{r_{1}}^{r_{1}} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}})</math>
:Now add <math>\Delta V_{AD}</math> to <math>\Delta V_{DC}</math> to get <math>\Delta V_{AC}</math>
:<math>\Delta V_{AC} = \Delta V_{AD} + \Delta V_{DC} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}}) + \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}}) = 0V </math>
:Notice that either path taken from A to C results in the same potential difference of 0V.

==Connectedness==
#How is this topic connected to something that you are interested in?
::I am very interested in electronics and circuitry so knowing that potential difference is path independent definitely helps in determining the voltage between 2 points in a complicated circuit with many paths.
#How is it connected to your major?
::I am an electrical engineering major so this knowledge is very applicable to determining the voltage between 2 points in a circuit that contains many loops or is very complex.

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.
Luigi Galvani, an Italian physicist, discovered something he called "animal electricity" which was two different metals connected in series with a frog leg and each other. However, Alessandro Volta, another Italian physicist, realized that the frog leg was acting as a conductor and a detector of electricity, or an electrolyte. He later replaced the frog leg with brine soaked paper and noticed the same results. Through this, Volta discovered electromotive force relating to a galvanic cell, realizing that the difference between the two electrode potentials caused electricity to flow.

== See also ==

https://en.wikipedia.org/wiki/Voltage

===Further reading===

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elewor.html#c2

===External links===

http://mathwiki.ucdavis.edu/Core/Calculus/Vector_Calculus/Integration_in_Vector_Fields/Path_Independence,_Conservative_Fields,_and_Potential_Functions

==References==

Matter & Interactions volume II by Ruth W. Chabay and Bruce A. Sherwood.

[[Category:Energy]]

File:Electrostatic definition of voltage.jpg

2016-11-28T01:14:32Z

Zwright8: Work is independent of the path in a static field.

Work is independent of the path in a static field.

Potential Difference Path Independence, claimed by Aditya Mohile

2016-11-27T23:26:31Z

Zwright8:

''Claimed by Aditya Mohile''

'''Claimed by Zachary Wright (Fall 2016)'''

Potential difference or Voltage is a scalar quantity that measures the difference of electric potential energy between two points. Potential energy differences depend only on final and initial states and is independent of the path taken to arrive at those states. Likewise, the potential difference between a final and initial state is independent of the path taken to arrive at those states, therefore the potential difference is said to be path independent. The units of measurement for potential difference is joules per coulomb, which is equal to volts.
==The Main Idea==

The potential difference between two points does not depend on the path taken to get from one point to another because the difference between the locations of the two points remains the same regardless of the path taken to get from one point to another.

===A Mathematical Model===
:<math>\Delta V_{AB} = \int_{A}^{B} \vec{E} \cdot d\vec{l} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>

The potential difference, <math>\Delta V_{AB}</math>, is the integral of the electrical field from point A to point B. This integral, however, is the dot product of <math>\vec{E}</math> and <math>d\vec{l}</math> which is equal to the sum of the negative electric field multiplied by the difference of the x,y and z components of the two points, as see from the equation above. This proves that the path taken from point A to point B does not matter, only the difference in the positions of points A and B are important in calculating the potential difference.

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===
[[File:Wiki_simple.jpg|center]]
'''Problem''': Locations A, B, and C are in a region of uniform electric field, as shown in the diagram above. Location A is at < -0.6, 0, 0> m. Location B is at < 0.4, 0, 0>. Location C is at < 0.4, -0.3, 0>m. In the region the electric field = < -750, 400, 0> N/C. Calculate the potential difference between A and C traveling from A to B then to C. Calculate the potential difference traveling from A to C directly.

'''Step 1'''
:Calculate the displacement vector from A to B.
:<math>\Delta l = < 0.4, 0, 0> - < -0.6, 0, 0> = < 1, 0, 0></math>
'''Step 2'''
:Calculate the potential difference between A and B.
:<math>\Delta V_{AB} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AB} = -(-750*1 + 400*0 + 0*0) = 750V</math>
'''Step 3'''
:Calculate the displacement vector from B to C.
:<math>\Delta l = < 0.4, -0.3, 0> - < 0.4, 0, 0> = < 0, -0.3, 0></math>
'''Step 4'''
:Calculate the potential difference between B and C.
:<math>\Delta V_{BC} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{BC} = -(-750*0 + 400*-0.3 + 0*0) = 120V</math>
'''Step 5'''
:Add <math>\Delta V_{AB}</math> to <math>\Delta V_{BC}</math> to get <math>\Delta V_{AC}</math>
:<math>\Delta V_{AC} = \Delta V_{AB} + \Delta V_{BC} = 750 + 120 = 870V</math>
Now you have to calculate the potential difference from A to C directly.

'''Step 1'''
:Calculate the displacement vector from A to C.
:<math>\Delta l = < 0.4, -0.3, 0> - < -0.6, 0, 0> = < 1, -0.3, 0></math>
'''Step 2'''
:Calculate the potential difference between A and C.
:<math>\Delta V_{AB} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AB} = -(-750*1 + 400*-0.3 + 0*0) = 750 + 120 = 870V</math>
As you can see, potential difference from A to C directly is the same as the potential difference from A to B then to C.

===Middling===
[[File:Wiki_middling.jpg|350px|center]]
'''Problem''': A capacitor consists of two charged disks of radius 4.7 m separated by a distance s = 2 mm (see the figure). The magnitude of the charge on each disk is 56 µC. Consider points A, B, C, and D inside the capacitor, as shown in the diagram. The distance s1 = 1.5 mm, and the distance s2 = 0.6 mm. (Assume the +x axis is to the right, the +y axis is up, and the +z axis is out.) Find the potential difference between B and D, traveling through A. Find the potential difference from B to D directly.
'''Step 1'''
:Calculate the electric field inside the capacitor.
:<math>\vec E_{cap} = \frac{Q/A}{\epsilon_{0}}</math>
:<math>\vec E_{cap} = \frac{(56*10^{-6})/(\pi*4.7^{2})}{8.85*10^{-12}} = < 91179.912, 0, 0></math>
The electric field only has an x-component because the field travels from the positively charged plate to the negatively charged plate, traveling in the +x direction.

'''Step 2'''
:Calculate the displacement vector from B to A.
:Set the origin at B.
:<math>\Delta l = < s_{1}, 0, 0> - < 0, 0, 0> = < s_{1}, 0, 0> = < 0.0015, 0, 0></math>
'''Step 3'''
:Calculate the potential difference between B and A.
:<math>\Delta V_{BA} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{BA} = -(91179.912*0.0015 + 0*0 + 0*0) = -136.770V</math>
'''Step 4'''
:Calculate the displacement vector from A to D
:<math>\Delta l = < s_{1}, s_{2}, 0> - < s_{1}, 0, 0> = < 0, s_{2}, 0> = < 0, s_{2}, 0> = < 0, 0.006, 0></math>
'''Step 5'''
:Calculate the potential difference between A and D.
:<math>\Delta V_{AD} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AD} = -(91179.912*0 + 0*0 + 0*0) = 0V</math>
'''Step 6'''
:Add <math>\Delta V_{BA}</math> to <math>\Delta V_{AD}</math> to get <math>\Delta V_{BD}</math>
:<math>\Delta V_{BD} = \Delta V_{BA} + \Delta V_{AD} = -136.770 + 0 = -136.770V</math>
Notice that the voltage is negative since we are going from the positively charged plate (high potential) to the negatively charged plate (low potential) losing potential energy in the process.

Now you have to calculate the potential difference from A to C directly.

'''Step 1'''
:Calculate the displacement vector from B to D.
:<math>\Delta l = < s_{1}, s_{2}, 0> - < 0, 0, 0> = < s_{1}, s_{2}, 0> = < s_{1}, s_{2}, 0> = < 0.0015, 0.006, 0></math>
'''Step 2'''
:Calculate the potential difference between B and D.
:<math>\Delta V_{AD} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AD} = -(91179.912*0.0015 + 0*0 + 0*0) = -136.770V</math>
Notice that potential difference traveling from B to A then D is the same as the potential difference traveling from B to D directly. This exemplifies that the potential difference does not depend on the path taken to get from one point to another.

===Difficult===
[[File:Wiki_difficult.png|center]]
'''Problem''': A uniform spherical shell of charge +Q is centered at point B, as shown in the figure below. Show that ΔV = VC − VA is independent of path by calculating ΔV for each of these two paths (actually do the integrals). (Use the following as necessary: Q, R for the radius of the spherical shell, r1 for the radius of the circular arc A → D → C with B at its center, and ε0.)

'''Step 1'''
:Calculate the potential difference from A to C following the path A → B → C.
:First, calculate the potential difference from A to B.
:<math>\Delta V_{AB} = \int_{A}^{B} \vec{E} \cdot d\vec{l}</math>
:In order to calculate the potential difference from A to B, calculate the electric field of the sphere.
:<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}}</math>
:Now plug the electric field calculated above into the integral and solve.
:<math>\Delta V_{AB} = \int_{A}^{B} \vec{E} \cdot d\vec{r} = \int_{r_{1}}^{R} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{R^{2}} - \frac{Q}{r_{1}^{2}})</math>
:Now calculate the potential difference from B to C.
:Use the electric field of the sphere calculated above to solve the below integral.
:<math>\Delta V_{BC} = \int_{B}^{C} \vec{E} \cdot d\vec{l}</math>
:<math>\Delta V_{BC} = \int_{R}^{r_{1}} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{R^{2}})</math>
:Now add <math>\Delta V_{AB}</math> to <math>\Delta V_{BC}</math> to get <math>\Delta V_{AC}</math>
:<math>\Delta V_{AC} = \Delta V_{AB} + \Delta V_{BC} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{R^{2}} - \frac{Q}{r_{1}^{2}}) + \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{R^{2}}) = 0V </math>
'''Step 2'''
:Calculate the potential difference from A to C following the path A → D → C.
:First calculate the potential difference from A to D.
:<math>\Delta V_{AD} = \int_{A}^{D} \vec{E} \cdot d\vec{l}</math>
:Use the same electric field of the sphere calculated in step 1, shown below.
:<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}}</math>
:Now plug the electric field calculated above into the integral and solve.
:<math>\Delta V_{AD} = \int_{A}^{D} \vec{E} \cdot d\vec{r} = \int_{r_{1}}^{r_{1}} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}}) = 0V</math>
:Now calculate the potential difference from D to C.
:Use the electric field of the sphere calculated above to solve the below integral.
:<math>\Delta V_{DC} = \int_{D}^{C} \vec{E} \cdot d\vec{l}</math>
:<math>\Delta V_{DC} = \int_{r_{1}}^{r_{1}} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}})</math>
:Now add <math>\Delta V_{AD}</math> to <math>\Delta V_{DC}</math> to get <math>\Delta V_{AC}</math>
:<math>\Delta V_{AC} = \Delta V_{AD} + \Delta V_{DC} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}}) + \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}}) = 0V </math>
:Notice that either path taken from A to C results in the same potential difference of 0V.

==Connectedness==
#How is this topic connected to something that you are interested in?
::I am very interested in electronics and circuitry so knowing that potential difference is path independent definitely helps in determining the voltage between 2 points in a complicated circuit with many paths.
#How is it connected to your major?
::I am an electrical engineering major so this knowledge is very applicable to determining the voltage between 2 points in a circuit that contains many loops or is very complex.

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.
Luigi Galvani, an Italian physicist, discovered something he called "animal electricity" which was two different metals connected in series with a frog leg and each other. However, Alessandro Volta, another Italian physicist, realized that the frog leg was acting as a conductor and a detector of electricity, or an electrolyte. He later replaced the frog leg with brine soaked paper and noticed the same results. Through this, Volta discovered electromotive force relating to a galvanic cell, realizing that the difference between the two electrode potentials caused electricity to flow.

== See also ==

https://en.wikipedia.org/wiki/Voltage

===Further reading===

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elewor.html#c2

===External links===

http://mathwiki.ucdavis.edu/Core/Calculus/Vector_Calculus/Integration_in_Vector_Fields/Path_Independence,_Conservative_Fields,_and_Potential_Functions

==References==

Matter & Interactions volume II by Ruth W. Chabay and Bruce A. Sherwood.

[[Category:Energy]]

Potential Difference Path Independence, claimed by Aditya Mohile

2016-11-27T23:25:05Z

Zwright8:

''Claimed by Aditya Mohile''

'''Claimed by Zachary Wright (Fall 2016)'''

Potential difference or Voltage is a scalar quantity that measures the difference of electric potential energy between two points. Potential energy differences depend only on final and initial states and is independent of the path taken to arrive at those states. Likewise, the potential difference between a final and initial state is independent of the path taken to arrive at those states, therefore the potential difference is said to be path independent. The units of measurement for potential difference is joules per coulomb, which is equal to volts.
==The Main Idea==

The potential difference between two points does not depend on the path taken to get from one point to another because the difference between the locations of the two points remains the same regardless of the path taken to get from one point to another.

===A Mathematical Model===
:<math>\Delta V_{AB} = \int_{A}^{B} \vec{E} \cdot d\vec{l} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>

The potential difference, <math>\Delta V_{AB}</math>, is the integral of the electrical field from point A to point B. This integral, however, is the dot product of <math>\vec{E}</math> and <math>d\vec{l}</math> which is equal to the sum of the negative electric field multiplied by the difference of the x,y and z components of the 2 points, as see from the equation above. This proves that the path taken from point A to point B does not matter, only the difference in the positions of points A and B are important in calculating the potential difference.

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===
[[File:Wiki_simple.jpg|center]]
'''Problem''': Locations A, B, and C are in a region of uniform electric field, as shown in the diagram above. Location A is at < -0.6, 0, 0> m. Location B is at < 0.4, 0, 0>. Location C is at < 0.4, -0.3, 0>m. In the region the electric field = < -750, 400, 0> N/C. Calculate the potential difference between A and C traveling from A to B then to C. Calculate the potential difference traveling from A to C directly.

'''Step 1'''
:Calculate the displacement vector from A to B.
:<math>\Delta l = < 0.4, 0, 0> - < -0.6, 0, 0> = < 1, 0, 0></math>
'''Step 2'''
:Calculate the potential difference between A and B.
:<math>\Delta V_{AB} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AB} = -(-750*1 + 400*0 + 0*0) = 750V</math>
'''Step 3'''
:Calculate the displacement vector from B to C.
:<math>\Delta l = < 0.4, -0.3, 0> - < 0.4, 0, 0> = < 0, -0.3, 0></math>
'''Step 4'''
:Calculate the potential difference between B and C.
:<math>\Delta V_{BC} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{BC} = -(-750*0 + 400*-0.3 + 0*0) = 120V</math>
'''Step 5'''
:Add <math>\Delta V_{AB}</math> to <math>\Delta V_{BC}</math> to get <math>\Delta V_{AC}</math>
:<math>\Delta V_{AC} = \Delta V_{AB} + \Delta V_{BC} = 750 + 120 = 870V</math>
Now you have to calculate the potential difference from A to C directly.

'''Step 1'''
:Calculate the displacement vector from A to C.
:<math>\Delta l = < 0.4, -0.3, 0> - < -0.6, 0, 0> = < 1, -0.3, 0></math>
'''Step 2'''
:Calculate the potential difference between A and C.
:<math>\Delta V_{AB} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AB} = -(-750*1 + 400*-0.3 + 0*0) = 750 + 120 = 870V</math>
As you can see, potential difference from A to C directly is the same as the potential difference from A to B then to C.

===Middling===
[[File:Wiki_middling.jpg|350px|center]]
'''Problem''': A capacitor consists of two charged disks of radius 4.7 m separated by a distance s = 2 mm (see the figure). The magnitude of the charge on each disk is 56 µC. Consider points A, B, C, and D inside the capacitor, as shown in the diagram. The distance s1 = 1.5 mm, and the distance s2 = 0.6 mm. (Assume the +x axis is to the right, the +y axis is up, and the +z axis is out.) Find the potential difference between B and D, traveling through A. Find the potential difference from B to D directly.
'''Step 1'''
:Calculate the electric field inside the capacitor.
:<math>\vec E_{cap} = \frac{Q/A}{\epsilon_{0}}</math>
:<math>\vec E_{cap} = \frac{(56*10^{-6})/(\pi*4.7^{2})}{8.85*10^{-12}} = < 91179.912, 0, 0></math>
The electric field only has an x-component because the field travels from the positively charged plate to the negatively charged plate, traveling in the +x direction.

'''Step 2'''
:Calculate the displacement vector from B to A.
:Set the origin at B.
:<math>\Delta l = < s_{1}, 0, 0> - < 0, 0, 0> = < s_{1}, 0, 0> = < 0.0015, 0, 0></math>
'''Step 3'''
:Calculate the potential difference between B and A.
:<math>\Delta V_{BA} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{BA} = -(91179.912*0.0015 + 0*0 + 0*0) = -136.770V</math>
'''Step 4'''
:Calculate the displacement vector from A to D
:<math>\Delta l = < s_{1}, s_{2}, 0> - < s_{1}, 0, 0> = < 0, s_{2}, 0> = < 0, s_{2}, 0> = < 0, 0.006, 0></math>
'''Step 5'''
:Calculate the potential difference between A and D.
:<math>\Delta V_{AD} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AD} = -(91179.912*0 + 0*0 + 0*0) = 0V</math>
'''Step 6'''
:Add <math>\Delta V_{BA}</math> to <math>\Delta V_{AD}</math> to get <math>\Delta V_{BD}</math>
:<math>\Delta V_{BD} = \Delta V_{BA} + \Delta V_{AD} = -136.770 + 0 = -136.770V</math>
Notice that the voltage is negative since we are going from the positively charged plate (high potential) to the negatively charged plate (low potential) losing potential energy in the process.

Now you have to calculate the potential difference from A to C directly.

'''Step 1'''
:Calculate the displacement vector from B to D.
:<math>\Delta l = < s_{1}, s_{2}, 0> - < 0, 0, 0> = < s_{1}, s_{2}, 0> = < s_{1}, s_{2}, 0> = < 0.0015, 0.006, 0></math>
'''Step 2'''
:Calculate the potential difference between B and D.
:<math>\Delta V_{AD} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AD} = -(91179.912*0.0015 + 0*0 + 0*0) = -136.770V</math>
Notice that potential difference traveling from B to A then D is the same as the potential difference traveling from B to D directly. This exemplifies that the potential difference does not depend on the path taken to get from one point to another.

===Difficult===
[[File:Wiki_difficult.png|center]]
'''Problem''': A uniform spherical shell of charge +Q is centered at point B, as shown in the figure below. Show that ΔV = VC − VA is independent of path by calculating ΔV for each of these two paths (actually do the integrals). (Use the following as necessary: Q, R for the radius of the spherical shell, r1 for the radius of the circular arc A → D → C with B at its center, and ε0.)

'''Step 1'''
:Calculate the potential difference from A to C following the path A → B → C.
:First, calculate the potential difference from A to B.
:<math>\Delta V_{AB} = \int_{A}^{B} \vec{E} \cdot d\vec{l}</math>
:In order to calculate the potential difference from A to B, calculate the electric field of the sphere.
:<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}}</math>
:Now plug the electric field calculated above into the integral and solve.
:<math>\Delta V_{AB} = \int_{A}^{B} \vec{E} \cdot d\vec{r} = \int_{r_{1}}^{R} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{R^{2}} - \frac{Q}{r_{1}^{2}})</math>
:Now calculate the potential difference from B to C.
:Use the electric field of the sphere calculated above to solve the below integral.
:<math>\Delta V_{BC} = \int_{B}^{C} \vec{E} \cdot d\vec{l}</math>
:<math>\Delta V_{BC} = \int_{R}^{r_{1}} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{R^{2}})</math>
:Now add <math>\Delta V_{AB}</math> to <math>\Delta V_{BC}</math> to get <math>\Delta V_{AC}</math>
:<math>\Delta V_{AC} = \Delta V_{AB} + \Delta V_{BC} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{R^{2}} - \frac{Q}{r_{1}^{2}}) + \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{R^{2}}) = 0V </math>
'''Step 2'''
:Calculate the potential difference from A to C following the path A → D → C.
:First calculate the potential difference from A to D.
:<math>\Delta V_{AD} = \int_{A}^{D} \vec{E} \cdot d\vec{l}</math>
:Use the same electric field of the sphere calculated in step 1, shown below.
:<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}}</math>
:Now plug the electric field calculated above into the integral and solve.
:<math>\Delta V_{AD} = \int_{A}^{D} \vec{E} \cdot d\vec{r} = \int_{r_{1}}^{r_{1}} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}}) = 0V</math>
:Now calculate the potential difference from D to C.
:Use the electric field of the sphere calculated above to solve the below integral.
:<math>\Delta V_{DC} = \int_{D}^{C} \vec{E} \cdot d\vec{l}</math>
:<math>\Delta V_{DC} = \int_{r_{1}}^{r_{1}} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}})</math>
:Now add <math>\Delta V_{AD}</math> to <math>\Delta V_{DC}</math> to get <math>\Delta V_{AC}</math>
:<math>\Delta V_{AC} = \Delta V_{AD} + \Delta V_{DC} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}}) + \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}}) = 0V </math>
:Notice that either path taken from A to C results in the same potential difference of 0V.

==Connectedness==
#How is this topic connected to something that you are interested in?
::I am very interested in electronics and circuitry so knowing that potential difference is path independent definitely helps in determining the voltage between 2 points in a complicated circuit with many paths.
#How is it connected to your major?
::I am an electrical engineering major so this knowledge is very applicable to determining the voltage between 2 points in a circuit that contains many loops or is very complex.

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.
Luigi Galvani, an Italian physicist, discovered something he called "animal electricity" which was two different metals connected in series with a frog leg and each other. However, Alessandro Volta, another Italian physicist, realized that the frog leg was acting as a conductor and a detector of electricity, or an electrolyte. He later replaced the frog leg with brine soaked paper and noticed the same results. Through this, Volta discovered electromotive force relating to a galvanic cell, realizing that the difference between the two electrode potentials caused electricity to flow.

== See also ==

https://en.wikipedia.org/wiki/Voltage

===Further reading===

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elewor.html#c2

===External links===

http://mathwiki.ucdavis.edu/Core/Calculus/Vector_Calculus/Integration_in_Vector_Fields/Path_Independence,_Conservative_Fields,_and_Potential_Functions

==References==

Matter & Interactions volume II by Ruth W. Chabay and Bruce A. Sherwood.

[[Category:Energy]]

Potential Difference Path Independence, claimed by Aditya Mohile

2016-11-27T22:51:47Z

Zwright8:

''Claimed by Aditya Mohile''

'''Claimed by Zachary Wright (Fall 2016)'''

Potential difference or Voltage is a scalar quantity that measures the difference of electric potential energy between two points. Potential energy differences depend only on final and initial states and is independent of the path taken to arrive at those states. Likewise, the potential difference between a final and initial state is independent of the path taken to arrive at those states, therefore the potential difference is said to be path independent.
==The Main Idea==

The potential difference between two points does not depend on the path taken to get from one point to another because the difference between the locations of the two points remains the same regardless of the path taken to get from one point to another.

===A Mathematical Model===
:<math>\Delta V_{AB} = \int_{A}^{B} \vec{E} \cdot d\vec{l} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>

The potential difference, <math>\Delta V_{AB}</math>, is the integral of the electrical field from point A to point B. This integral, however, is the dot product of <math>\vec{E}</math> and <math>d\vec{l}</math> which is equal to the sum of the negative electric field multiplied by the difference of the x,y and z components of the 2 points, as see from the equation above. This proves that the path taken from point A to point B does not matter, only the difference in the positions of points A and B are important in calculating the potential difference.

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===
[[File:Wiki_simple.jpg|center]]
'''Problem''': Locations A, B, and C are in a region of uniform electric field, as shown in the diagram above. Location A is at < -0.6, 0, 0> m. Location B is at < 0.4, 0, 0>. Location C is at < 0.4, -0.3, 0>m. In the region the electric field = < -750, 400, 0> N/C. Calculate the potential difference between A and C traveling from A to B then to C. Calculate the potential difference traveling from A to C directly.

'''Step 1'''
:Calculate the displacement vector from A to B.
:<math>\Delta l = < 0.4, 0, 0> - < -0.6, 0, 0> = < 1, 0, 0></math>
'''Step 2'''
:Calculate the potential difference between A and B.
:<math>\Delta V_{AB} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AB} = -(-750*1 + 400*0 + 0*0) = 750V</math>
'''Step 3'''
:Calculate the displacement vector from B to C.
:<math>\Delta l = < 0.4, -0.3, 0> - < 0.4, 0, 0> = < 0, -0.3, 0></math>
'''Step 4'''
:Calculate the potential difference between B and C.
:<math>\Delta V_{BC} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{BC} = -(-750*0 + 400*-0.3 + 0*0) = 120V</math>
'''Step 5'''
:Add <math>\Delta V_{AB}</math> to <math>\Delta V_{BC}</math> to get <math>\Delta V_{AC}</math>
:<math>\Delta V_{AC} = \Delta V_{AB} + \Delta V_{BC} = 750 + 120 = 870V</math>
Now you have to calculate the potential difference from A to C directly.

'''Step 1'''
:Calculate the displacement vector from A to C.
:<math>\Delta l = < 0.4, -0.3, 0> - < -0.6, 0, 0> = < 1, -0.3, 0></math>
'''Step 2'''
:Calculate the potential difference between A and C.
:<math>\Delta V_{AB} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AB} = -(-750*1 + 400*-0.3 + 0*0) = 750 + 120 = 870V</math>
As you can see, potential difference from A to C directly is the same as the potential difference from A to B then to C.

===Middling===
[[File:Wiki_middling.jpg|350px|center]]
'''Problem''': A capacitor consists of two charged disks of radius 4.7 m separated by a distance s = 2 mm (see the figure). The magnitude of the charge on each disk is 56 µC. Consider points A, B, C, and D inside the capacitor, as shown in the diagram. The distance s1 = 1.5 mm, and the distance s2 = 0.6 mm. (Assume the +x axis is to the right, the +y axis is up, and the +z axis is out.) Find the potential difference between B and D, traveling through A. Find the potential difference from B to D directly.
'''Step 1'''
:Calculate the electric field inside the capacitor.
:<math>\vec E_{cap} = \frac{Q/A}{\epsilon_{0}}</math>
:<math>\vec E_{cap} = \frac{(56*10^{-6})/(\pi*4.7^{2})}{8.85*10^{-12}} = < 91179.912, 0, 0></math>
The electric field only has an x-component because the field travels from the positively charged plate to the negatively charged plate, traveling in the +x direction.

'''Step 2'''
:Calculate the displacement vector from B to A.
:Set the origin at B.
:<math>\Delta l = < s_{1}, 0, 0> - < 0, 0, 0> = < s_{1}, 0, 0> = < 0.0015, 0, 0></math>
'''Step 3'''
:Calculate the potential difference between B and A.
:<math>\Delta V_{BA} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{BA} = -(91179.912*0.0015 + 0*0 + 0*0) = -136.770V</math>
'''Step 4'''
:Calculate the displacement vector from A to D
:<math>\Delta l = < s_{1}, s_{2}, 0> - < s_{1}, 0, 0> = < 0, s_{2}, 0> = < 0, s_{2}, 0> = < 0, 0.006, 0></math>
'''Step 5'''
:Calculate the potential difference between A and D.
:<math>\Delta V_{AD} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AD} = -(91179.912*0 + 0*0 + 0*0) = 0V</math>
'''Step 6'''
:Add <math>\Delta V_{BA}</math> to <math>\Delta V_{AD}</math> to get <math>\Delta V_{BD}</math>
:<math>\Delta V_{BD} = \Delta V_{BA} + \Delta V_{AD} = -136.770 + 0 = -136.770V</math>
Notice that the voltage is negative since we are going from the positively charged plate (high potential) to the negatively charged plate (low potential) losing potential energy in the process.

Now you have to calculate the potential difference from A to C directly.

'''Step 1'''
:Calculate the displacement vector from B to D.
:<math>\Delta l = < s_{1}, s_{2}, 0> - < 0, 0, 0> = < s_{1}, s_{2}, 0> = < s_{1}, s_{2}, 0> = < 0.0015, 0.006, 0></math>
'''Step 2'''
:Calculate the potential difference between B and D.
:<math>\Delta V_{AD} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AD} = -(91179.912*0.0015 + 0*0 + 0*0) = -136.770V</math>
Notice that potential difference traveling from B to A then D is the same as the potential difference traveling from B to D directly. This exemplifies that the potential difference does not depend on the path taken to get from one point to another.

===Difficult===
[[File:Wiki_difficult.png|center]]
'''Problem''': A uniform spherical shell of charge +Q is centered at point B, as shown in the figure below. Show that ΔV = VC − VA is independent of path by calculating ΔV for each of these two paths (actually do the integrals). (Use the following as necessary: Q, R for the radius of the spherical shell, r1 for the radius of the circular arc A → D → C with B at its center, and ε0.)

'''Step 1'''
:Calculate the potential difference from A to C following the path A → B → C.
:First, calculate the potential difference from A to B.
:<math>\Delta V_{AB} = \int_{A}^{B} \vec{E} \cdot d\vec{l}</math>
:In order to calculate the potential difference from A to B, calculate the electric field of the sphere.
:<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}}</math>
:Now plug the electric field calculated above into the integral and solve.
:<math>\Delta V_{AB} = \int_{A}^{B} \vec{E} \cdot d\vec{r} = \int_{r_{1}}^{R} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{R^{2}} - \frac{Q}{r_{1}^{2}})</math>
:Now calculate the potential difference from B to C.
:Use the electric field of the sphere calculated above to solve the below integral.
:<math>\Delta V_{BC} = \int_{B}^{C} \vec{E} \cdot d\vec{l}</math>
:<math>\Delta V_{BC} = \int_{R}^{r_{1}} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{R^{2}})</math>
:Now add <math>\Delta V_{AB}</math> to <math>\Delta V_{BC}</math> to get <math>\Delta V_{AC}</math>
:<math>\Delta V_{AC} = \Delta V_{AB} + \Delta V_{BC} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{R^{2}} - \frac{Q}{r_{1}^{2}}) + \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{R^{2}}) = 0V </math>
'''Step 2'''
:Calculate the potential difference from A to C following the path A → D → C.
:First calculate the potential difference from A to D.
:<math>\Delta V_{AD} = \int_{A}^{D} \vec{E} \cdot d\vec{l}</math>
:Use the same electric field of the sphere calculated in step 1, shown below.
:<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}}</math>
:Now plug the electric field calculated above into the integral and solve.
:<math>\Delta V_{AD} = \int_{A}^{D} \vec{E} \cdot d\vec{r} = \int_{r_{1}}^{r_{1}} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}}) = 0V</math>
:Now calculate the potential difference from D to C.
:Use the electric field of the sphere calculated above to solve the below integral.
:<math>\Delta V_{DC} = \int_{D}^{C} \vec{E} \cdot d\vec{l}</math>
:<math>\Delta V_{DC} = \int_{r_{1}}^{r_{1}} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}})</math>
:Now add <math>\Delta V_{AD}</math> to <math>\Delta V_{DC}</math> to get <math>\Delta V_{AC}</math>
:<math>\Delta V_{AC} = \Delta V_{AD} + \Delta V_{DC} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}}) + \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}}) = 0V </math>
:Notice that either path taken from A to C results in the same potential difference of 0V.

==Connectedness==
#How is this topic connected to something that you are interested in?
::I am very interested in electronics and circuitry so knowing that potential difference is path independent definitely helps in determining the voltage between 2 points in a complicated circuit with many paths.
#How is it connected to your major?
::I am an electrical engineering major so this knowledge is very applicable to determining the voltage between 2 points in a circuit that contains many loops or is very complex.

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.
Luigi Galvani, an Italian physicist, discovered something he called "animal electricity" which was two different metals connected in series with a frog leg and each other. However, Alessandro Volta, another Italian physicist, realized that the frog leg was acting as a conductor and a detector of electricity, or an electrolyte. He later replaced the frog leg with brine soaked paper and noticed the same results. Through this, Volta discovered electromotive force relating to a galvanic cell, realizing that the difference between the two electrode potentials caused electricity to flow.

== See also ==

https://en.wikipedia.org/wiki/Voltage

===Further reading===

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elewor.html#c2

===External links===

http://mathwiki.ucdavis.edu/Core/Calculus/Vector_Calculus/Integration_in_Vector_Fields/Path_Independence,_Conservative_Fields,_and_Potential_Functions

==References==

Matter & Interactions volume II by Ruth W. Chabay and Bruce A. Sherwood.

[[Category:Energy]]

Potential Difference Path Independence, claimed by Aditya Mohile

2016-11-27T22:51:04Z

Zwright8:

''Claimed by Aditya Mohile''

'''Claimed by Zachary Wright (Fall 2016)'''

Potential difference or Voltage is a scalar quantity that measures the difference of electric potential energy between two points. Potential energy differences depend only on final and initial states and is independent of the path taken to arrive at those final and initial states. Likewise, the potential difference between a final and initial state is independent of the path taken to arrive at those states, therefore the potential difference is said to be path independent.
==The Main Idea==

The potential difference between two points does not depend on the path taken to get from one point to another because the difference between the locations of the two points remains the same regardless of the path taken to get from one point to another.

===A Mathematical Model===
:<math>\Delta V_{AB} = \int_{A}^{B} \vec{E} \cdot d\vec{l} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>

The potential difference, <math>\Delta V_{AB}</math>, is the integral of the electrical field from point A to point B. This integral, however, is the dot product of <math>\vec{E}</math> and <math>d\vec{l}</math> which is equal to the sum of the negative electric field multiplied by the difference of the x,y and z components of the 2 points, as see from the equation above. This proves that the path taken from point A to point B does not matter, only the difference in the positions of points A and B are important in calculating the potential difference.

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===
[[File:Wiki_simple.jpg|center]]
'''Problem''': Locations A, B, and C are in a region of uniform electric field, as shown in the diagram above. Location A is at < -0.6, 0, 0> m. Location B is at < 0.4, 0, 0>. Location C is at < 0.4, -0.3, 0>m. In the region the electric field = < -750, 400, 0> N/C. Calculate the potential difference between A and C traveling from A to B then to C. Calculate the potential difference traveling from A to C directly.

'''Step 1'''
:Calculate the displacement vector from A to B.
:<math>\Delta l = < 0.4, 0, 0> - < -0.6, 0, 0> = < 1, 0, 0></math>
'''Step 2'''
:Calculate the potential difference between A and B.
:<math>\Delta V_{AB} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AB} = -(-750*1 + 400*0 + 0*0) = 750V</math>
'''Step 3'''
:Calculate the displacement vector from B to C.
:<math>\Delta l = < 0.4, -0.3, 0> - < 0.4, 0, 0> = < 0, -0.3, 0></math>
'''Step 4'''
:Calculate the potential difference between B and C.
:<math>\Delta V_{BC} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{BC} = -(-750*0 + 400*-0.3 + 0*0) = 120V</math>
'''Step 5'''
:Add <math>\Delta V_{AB}</math> to <math>\Delta V_{BC}</math> to get <math>\Delta V_{AC}</math>
:<math>\Delta V_{AC} = \Delta V_{AB} + \Delta V_{BC} = 750 + 120 = 870V</math>
Now you have to calculate the potential difference from A to C directly.

'''Step 1'''
:Calculate the displacement vector from A to C.
:<math>\Delta l = < 0.4, -0.3, 0> - < -0.6, 0, 0> = < 1, -0.3, 0></math>
'''Step 2'''
:Calculate the potential difference between A and C.
:<math>\Delta V_{AB} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AB} = -(-750*1 + 400*-0.3 + 0*0) = 750 + 120 = 870V</math>
As you can see, potential difference from A to C directly is the same as the potential difference from A to B then to C.

===Middling===
[[File:Wiki_middling.jpg|350px|center]]
'''Problem''': A capacitor consists of two charged disks of radius 4.7 m separated by a distance s = 2 mm (see the figure). The magnitude of the charge on each disk is 56 µC. Consider points A, B, C, and D inside the capacitor, as shown in the diagram. The distance s1 = 1.5 mm, and the distance s2 = 0.6 mm. (Assume the +x axis is to the right, the +y axis is up, and the +z axis is out.) Find the potential difference between B and D, traveling through A. Find the potential difference from B to D directly.
'''Step 1'''
:Calculate the electric field inside the capacitor.
:<math>\vec E_{cap} = \frac{Q/A}{\epsilon_{0}}</math>
:<math>\vec E_{cap} = \frac{(56*10^{-6})/(\pi*4.7^{2})}{8.85*10^{-12}} = < 91179.912, 0, 0></math>
The electric field only has an x-component because the field travels from the positively charged plate to the negatively charged plate, traveling in the +x direction.

'''Step 2'''
:Calculate the displacement vector from B to A.
:Set the origin at B.
:<math>\Delta l = < s_{1}, 0, 0> - < 0, 0, 0> = < s_{1}, 0, 0> = < 0.0015, 0, 0></math>
'''Step 3'''
:Calculate the potential difference between B and A.
:<math>\Delta V_{BA} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{BA} = -(91179.912*0.0015 + 0*0 + 0*0) = -136.770V</math>
'''Step 4'''
:Calculate the displacement vector from A to D
:<math>\Delta l = < s_{1}, s_{2}, 0> - < s_{1}, 0, 0> = < 0, s_{2}, 0> = < 0, s_{2}, 0> = < 0, 0.006, 0></math>
'''Step 5'''
:Calculate the potential difference between A and D.
:<math>\Delta V_{AD} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AD} = -(91179.912*0 + 0*0 + 0*0) = 0V</math>
'''Step 6'''
:Add <math>\Delta V_{BA}</math> to <math>\Delta V_{AD}</math> to get <math>\Delta V_{BD}</math>
:<math>\Delta V_{BD} = \Delta V_{BA} + \Delta V_{AD} = -136.770 + 0 = -136.770V</math>
Notice that the voltage is negative since we are going from the positively charged plate (high potential) to the negatively charged plate (low potential) losing potential energy in the process.

Now you have to calculate the potential difference from A to C directly.

'''Step 1'''
:Calculate the displacement vector from B to D.
:<math>\Delta l = < s_{1}, s_{2}, 0> - < 0, 0, 0> = < s_{1}, s_{2}, 0> = < s_{1}, s_{2}, 0> = < 0.0015, 0.006, 0></math>
'''Step 2'''
:Calculate the potential difference between B and D.
:<math>\Delta V_{AD} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AD} = -(91179.912*0.0015 + 0*0 + 0*0) = -136.770V</math>
Notice that potential difference traveling from B to A then D is the same as the potential difference traveling from B to D directly. This exemplifies that the potential difference does not depend on the path taken to get from one point to another.

===Difficult===
[[File:Wiki_difficult.png|center]]
'''Problem''': A uniform spherical shell of charge +Q is centered at point B, as shown in the figure below. Show that ΔV = VC − VA is independent of path by calculating ΔV for each of these two paths (actually do the integrals). (Use the following as necessary: Q, R for the radius of the spherical shell, r1 for the radius of the circular arc A → D → C with B at its center, and ε0.)

'''Step 1'''
:Calculate the potential difference from A to C following the path A → B → C.
:First, calculate the potential difference from A to B.
:<math>\Delta V_{AB} = \int_{A}^{B} \vec{E} \cdot d\vec{l}</math>
:In order to calculate the potential difference from A to B, calculate the electric field of the sphere.
:<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}}</math>
:Now plug the electric field calculated above into the integral and solve.
:<math>\Delta V_{AB} = \int_{A}^{B} \vec{E} \cdot d\vec{r} = \int_{r_{1}}^{R} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{R^{2}} - \frac{Q}{r_{1}^{2}})</math>
:Now calculate the potential difference from B to C.
:Use the electric field of the sphere calculated above to solve the below integral.
:<math>\Delta V_{BC} = \int_{B}^{C} \vec{E} \cdot d\vec{l}</math>
:<math>\Delta V_{BC} = \int_{R}^{r_{1}} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{R^{2}})</math>
:Now add <math>\Delta V_{AB}</math> to <math>\Delta V_{BC}</math> to get <math>\Delta V_{AC}</math>
:<math>\Delta V_{AC} = \Delta V_{AB} + \Delta V_{BC} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{R^{2}} - \frac{Q}{r_{1}^{2}}) + \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{R^{2}}) = 0V </math>
'''Step 2'''
:Calculate the potential difference from A to C following the path A → D → C.
:First calculate the potential difference from A to D.
:<math>\Delta V_{AD} = \int_{A}^{D} \vec{E} \cdot d\vec{l}</math>
:Use the same electric field of the sphere calculated in step 1, shown below.
:<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}}</math>
:Now plug the electric field calculated above into the integral and solve.
:<math>\Delta V_{AD} = \int_{A}^{D} \vec{E} \cdot d\vec{r} = \int_{r_{1}}^{r_{1}} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}}) = 0V</math>
:Now calculate the potential difference from D to C.
:Use the electric field of the sphere calculated above to solve the below integral.
:<math>\Delta V_{DC} = \int_{D}^{C} \vec{E} \cdot d\vec{l}</math>
:<math>\Delta V_{DC} = \int_{r_{1}}^{r_{1}} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}})</math>
:Now add <math>\Delta V_{AD}</math> to <math>\Delta V_{DC}</math> to get <math>\Delta V_{AC}</math>
:<math>\Delta V_{AC} = \Delta V_{AD} + \Delta V_{DC} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}}) + \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}}) = 0V </math>
:Notice that either path taken from A to C results in the same potential difference of 0V.

==Connectedness==
#How is this topic connected to something that you are interested in?
::I am very interested in electronics and circuitry so knowing that potential difference is path independent definitely helps in determining the voltage between 2 points in a complicated circuit with many paths.
#How is it connected to your major?
::I am an electrical engineering major so this knowledge is very applicable to determining the voltage between 2 points in a circuit that contains many loops or is very complex.

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.
Luigi Galvani, an Italian physicist, discovered something he called "animal electricity" which was two different metals connected in series with a frog leg and each other. However, Alessandro Volta, another Italian physicist, realized that the frog leg was acting as a conductor and a detector of electricity, or an electrolyte. He later replaced the frog leg with brine soaked paper and noticed the same results. Through this, Volta discovered electromotive force relating to a galvanic cell, realizing that the difference between the two electrode potentials caused electricity to flow.

== See also ==

https://en.wikipedia.org/wiki/Voltage

===Further reading===

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elewor.html#c2

===External links===

http://mathwiki.ucdavis.edu/Core/Calculus/Vector_Calculus/Integration_in_Vector_Fields/Path_Independence,_Conservative_Fields,_and_Potential_Functions

==References==

Matter & Interactions volume II by Ruth W. Chabay and Bruce A. Sherwood.

[[Category:Energy]]

Potential Difference Path Independence, claimed by Aditya Mohile

2016-11-27T22:50:16Z

Zwright8:

''Claimed by Aditya Mohile''

'''Claimed by Zachary Wright (Fall 2016)'''

Potential difference or Voltage is a scalar quantity that measures the difference of electric potential between 2 points. Potential energy differences depend only on final and initial states and is independent of the path taken to arrive at those final and initial states. Likewise, the potential difference between a final and initial state is independent of the path taken to arrive at those states, therefore the potential difference is said to be path independent.
==The Main Idea==

The potential difference between two points does not depend on the path taken to get from one point to another because the difference between the locations of the two points remains the same regardless of the path taken to get from one point to another.

===A Mathematical Model===
:<math>\Delta V_{AB} = \int_{A}^{B} \vec{E} \cdot d\vec{l} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>

The potential difference, <math>\Delta V_{AB}</math>, is the integral of the electrical field from point A to point B. This integral, however, is the dot product of <math>\vec{E}</math> and <math>d\vec{l}</math> which is equal to the sum of the negative electric field multiplied by the difference of the x,y and z components of the 2 points, as see from the equation above. This proves that the path taken from point A to point B does not matter, only the difference in the positions of points A and B are important in calculating the potential difference.

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===
[[File:Wiki_simple.jpg|center]]
'''Problem''': Locations A, B, and C are in a region of uniform electric field, as shown in the diagram above. Location A is at < -0.6, 0, 0> m. Location B is at < 0.4, 0, 0>. Location C is at < 0.4, -0.3, 0>m. In the region the electric field = < -750, 400, 0> N/C. Calculate the potential difference between A and C traveling from A to B then to C. Calculate the potential difference traveling from A to C directly.

'''Step 1'''
:Calculate the displacement vector from A to B.
:<math>\Delta l = < 0.4, 0, 0> - < -0.6, 0, 0> = < 1, 0, 0></math>
'''Step 2'''
:Calculate the potential difference between A and B.
:<math>\Delta V_{AB} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AB} = -(-750*1 + 400*0 + 0*0) = 750V</math>
'''Step 3'''
:Calculate the displacement vector from B to C.
:<math>\Delta l = < 0.4, -0.3, 0> - < 0.4, 0, 0> = < 0, -0.3, 0></math>
'''Step 4'''
:Calculate the potential difference between B and C.
:<math>\Delta V_{BC} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{BC} = -(-750*0 + 400*-0.3 + 0*0) = 120V</math>
'''Step 5'''
:Add <math>\Delta V_{AB}</math> to <math>\Delta V_{BC}</math> to get <math>\Delta V_{AC}</math>
:<math>\Delta V_{AC} = \Delta V_{AB} + \Delta V_{BC} = 750 + 120 = 870V</math>
Now you have to calculate the potential difference from A to C directly.

'''Step 1'''
:Calculate the displacement vector from A to C.
:<math>\Delta l = < 0.4, -0.3, 0> - < -0.6, 0, 0> = < 1, -0.3, 0></math>
'''Step 2'''
:Calculate the potential difference between A and C.
:<math>\Delta V_{AB} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AB} = -(-750*1 + 400*-0.3 + 0*0) = 750 + 120 = 870V</math>
As you can see, potential difference from A to C directly is the same as the potential difference from A to B then to C.

===Middling===
[[File:Wiki_middling.jpg|350px|center]]
'''Problem''': A capacitor consists of two charged disks of radius 4.7 m separated by a distance s = 2 mm (see the figure). The magnitude of the charge on each disk is 56 µC. Consider points A, B, C, and D inside the capacitor, as shown in the diagram. The distance s1 = 1.5 mm, and the distance s2 = 0.6 mm. (Assume the +x axis is to the right, the +y axis is up, and the +z axis is out.) Find the potential difference between B and D, traveling through A. Find the potential difference from B to D directly.
'''Step 1'''
:Calculate the electric field inside the capacitor.
:<math>\vec E_{cap} = \frac{Q/A}{\epsilon_{0}}</math>
:<math>\vec E_{cap} = \frac{(56*10^{-6})/(\pi*4.7^{2})}{8.85*10^{-12}} = < 91179.912, 0, 0></math>
The electric field only has an x-component because the field travels from the positively charged plate to the negatively charged plate, traveling in the +x direction.

'''Step 2'''
:Calculate the displacement vector from B to A.
:Set the origin at B.
:<math>\Delta l = < s_{1}, 0, 0> - < 0, 0, 0> = < s_{1}, 0, 0> = < 0.0015, 0, 0></math>
'''Step 3'''
:Calculate the potential difference between B and A.
:<math>\Delta V_{BA} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{BA} = -(91179.912*0.0015 + 0*0 + 0*0) = -136.770V</math>
'''Step 4'''
:Calculate the displacement vector from A to D
:<math>\Delta l = < s_{1}, s_{2}, 0> - < s_{1}, 0, 0> = < 0, s_{2}, 0> = < 0, s_{2}, 0> = < 0, 0.006, 0></math>
'''Step 5'''
:Calculate the potential difference between A and D.
:<math>\Delta V_{AD} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AD} = -(91179.912*0 + 0*0 + 0*0) = 0V</math>
'''Step 6'''
:Add <math>\Delta V_{BA}</math> to <math>\Delta V_{AD}</math> to get <math>\Delta V_{BD}</math>
:<math>\Delta V_{BD} = \Delta V_{BA} + \Delta V_{AD} = -136.770 + 0 = -136.770V</math>
Notice that the voltage is negative since we are going from the positively charged plate (high potential) to the negatively charged plate (low potential) losing potential energy in the process.

Now you have to calculate the potential difference from A to C directly.

'''Step 1'''
:Calculate the displacement vector from B to D.
:<math>\Delta l = < s_{1}, s_{2}, 0> - < 0, 0, 0> = < s_{1}, s_{2}, 0> = < s_{1}, s_{2}, 0> = < 0.0015, 0.006, 0></math>
'''Step 2'''
:Calculate the potential difference between B and D.
:<math>\Delta V_{AD} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AD} = -(91179.912*0.0015 + 0*0 + 0*0) = -136.770V</math>
Notice that potential difference traveling from B to A then D is the same as the potential difference traveling from B to D directly. This exemplifies that the potential difference does not depend on the path taken to get from one point to another.

===Difficult===
[[File:Wiki_difficult.png|center]]
'''Problem''': A uniform spherical shell of charge +Q is centered at point B, as shown in the figure below. Show that ΔV = VC − VA is independent of path by calculating ΔV for each of these two paths (actually do the integrals). (Use the following as necessary: Q, R for the radius of the spherical shell, r1 for the radius of the circular arc A → D → C with B at its center, and ε0.)

'''Step 1'''
:Calculate the potential difference from A to C following the path A → B → C.
:First, calculate the potential difference from A to B.
:<math>\Delta V_{AB} = \int_{A}^{B} \vec{E} \cdot d\vec{l}</math>
:In order to calculate the potential difference from A to B, calculate the electric field of the sphere.
:<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}}</math>
:Now plug the electric field calculated above into the integral and solve.
:<math>\Delta V_{AB} = \int_{A}^{B} \vec{E} \cdot d\vec{r} = \int_{r_{1}}^{R} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{R^{2}} - \frac{Q}{r_{1}^{2}})</math>
:Now calculate the potential difference from B to C.
:Use the electric field of the sphere calculated above to solve the below integral.
:<math>\Delta V_{BC} = \int_{B}^{C} \vec{E} \cdot d\vec{l}</math>
:<math>\Delta V_{BC} = \int_{R}^{r_{1}} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{R^{2}})</math>
:Now add <math>\Delta V_{AB}</math> to <math>\Delta V_{BC}</math> to get <math>\Delta V_{AC}</math>
:<math>\Delta V_{AC} = \Delta V_{AB} + \Delta V_{BC} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{R^{2}} - \frac{Q}{r_{1}^{2}}) + \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{R^{2}}) = 0V </math>
'''Step 2'''
:Calculate the potential difference from A to C following the path A → D → C.
:First calculate the potential difference from A to D.
:<math>\Delta V_{AD} = \int_{A}^{D} \vec{E} \cdot d\vec{l}</math>
:Use the same electric field of the sphere calculated in step 1, shown below.
:<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}}</math>
:Now plug the electric field calculated above into the integral and solve.
:<math>\Delta V_{AD} = \int_{A}^{D} \vec{E} \cdot d\vec{r} = \int_{r_{1}}^{r_{1}} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}}) = 0V</math>
:Now calculate the potential difference from D to C.
:Use the electric field of the sphere calculated above to solve the below integral.
:<math>\Delta V_{DC} = \int_{D}^{C} \vec{E} \cdot d\vec{l}</math>
:<math>\Delta V_{DC} = \int_{r_{1}}^{r_{1}} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}})</math>
:Now add <math>\Delta V_{AD}</math> to <math>\Delta V_{DC}</math> to get <math>\Delta V_{AC}</math>
:<math>\Delta V_{AC} = \Delta V_{AD} + \Delta V_{DC} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}}) + \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}}) = 0V </math>
:Notice that either path taken from A to C results in the same potential difference of 0V.

==Connectedness==
#How is this topic connected to something that you are interested in?
::I am very interested in electronics and circuitry so knowing that potential difference is path independent definitely helps in determining the voltage between 2 points in a complicated circuit with many paths.
#How is it connected to your major?
::I am an electrical engineering major so this knowledge is very applicable to determining the voltage between 2 points in a circuit that contains many loops or is very complex.

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.
Luigi Galvani, an Italian physicist, discovered something he called "animal electricity" which was two different metals connected in series with a frog leg and each other. However, Alessandro Volta, another Italian physicist, realized that the frog leg was acting as a conductor and a detector of electricity, or an electrolyte. He later replaced the frog leg with brine soaked paper and noticed the same results. Through this, Volta discovered electromotive force relating to a galvanic cell, realizing that the difference between the two electrode potentials caused electricity to flow.

== See also ==

https://en.wikipedia.org/wiki/Voltage

===Further reading===

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elewor.html#c2

===External links===

http://mathwiki.ucdavis.edu/Core/Calculus/Vector_Calculus/Integration_in_Vector_Fields/Path_Independence,_Conservative_Fields,_and_Potential_Functions

==References==

Matter & Interactions volume II by Ruth W. Chabay and Bruce A. Sherwood.

[[Category:Energy]]

Potential Difference Path Independence, claimed by Aditya Mohile

2016-11-27T21:33:52Z

Zwright8:

''Claimed by Aditya Mohile''

'''Claimed by Zachary Wright (Fall 2016)'''

Potential difference is a scalar quantity that measures the difference of electric potential between 2 points. The path taken from one point to another does not affect the potential difference therefore potential difference is said to be path independent.
==The Main Idea==

The potential difference between two points does not depend on the path taken to get from one point to another because the difference between the locations of the two points remain the same regardless of the path taken to get from one to another.

===A Mathematical Model===
:<math>\Delta V_{AB} = \int_{A}^{B} \vec{E} \cdot d\vec{l} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>

The potential difference, <math>\Delta V_{AB}</math>, is the integral of the electrical field from point A to point B. This integral, however, is the dot product of <math>\vec{E}</math> and <math>d\vec{l}</math> which is equal to the sum of the negative electric field multiplied by the difference of the x,y and z components of the 2 points, as see from the equation above. This proves that the path taken from point A to point B does not matter, only the difference in the positions of points A and B are important in calculating the potential difference.

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===
[[File:Wiki_simple.jpg|center]]
'''Problem''': Locations A, B, and C are in a region of uniform electric field, as shown in the diagram above. Location A is at < -0.6, 0, 0> m. Location B is at < 0.4, 0, 0>. Location C is at < 0.4, -0.3, 0>m. In the region the electric field = < -750, 400, 0> N/C. Calculate the potential difference between A and C traveling from A to B then to C. Calculate the potential difference traveling from A to C directly.

'''Step 1'''
:Calculate the displacement vector from A to B.
:<math>\Delta l = < 0.4, 0, 0> - < -0.6, 0, 0> = < 1, 0, 0></math>
'''Step 2'''
:Calculate the potential difference between A and B.
:<math>\Delta V_{AB} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AB} = -(-750*1 + 400*0 + 0*0) = 750V</math>
'''Step 3'''
:Calculate the displacement vector from B to C.
:<math>\Delta l = < 0.4, -0.3, 0> - < 0.4, 0, 0> = < 0, -0.3, 0></math>
'''Step 4'''
:Calculate the potential difference between B and C.
:<math>\Delta V_{BC} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{BC} = -(-750*0 + 400*-0.3 + 0*0) = 120V</math>
'''Step 5'''
:Add <math>\Delta V_{AB}</math> to <math>\Delta V_{BC}</math> to get <math>\Delta V_{AC}</math>
:<math>\Delta V_{AC} = \Delta V_{AB} + \Delta V_{BC} = 750 + 120 = 870V</math>
Now you have to calculate the potential difference from A to C directly.

'''Step 1'''
:Calculate the displacement vector from A to C.
:<math>\Delta l = < 0.4, -0.3, 0> - < -0.6, 0, 0> = < 1, -0.3, 0></math>
'''Step 2'''
:Calculate the potential difference between A and C.
:<math>\Delta V_{AB} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AB} = -(-750*1 + 400*-0.3 + 0*0) = 750 + 120 = 870V</math>
As you can see, potential difference from A to C directly is the same as the potential difference from A to B then to C.

===Middling===
[[File:Wiki_middling.jpg|350px|center]]
'''Problem''': A capacitor consists of two charged disks of radius 4.7 m separated by a distance s = 2 mm (see the figure). The magnitude of the charge on each disk is 56 µC. Consider points A, B, C, and D inside the capacitor, as shown in the diagram. The distance s1 = 1.5 mm, and the distance s2 = 0.6 mm. (Assume the +x axis is to the right, the +y axis is up, and the +z axis is out.) Find the potential difference between B and D, traveling through A. Find the potential difference from B to D directly.
'''Step 1'''
:Calculate the electric field inside the capacitor.
:<math>\vec E_{cap} = \frac{Q/A}{\epsilon_{0}}</math>
:<math>\vec E_{cap} = \frac{(56*10^{-6})/(\pi*4.7^{2})}{8.85*10^{-12}} = < 91179.912, 0, 0></math>
The electric field only has an x-component because the field travels from the positively charged plate to the negatively charged plate, traveling in the +x direction.

'''Step 2'''
:Calculate the displacement vector from B to A.
:Set the origin at B.
:<math>\Delta l = < s_{1}, 0, 0> - < 0, 0, 0> = < s_{1}, 0, 0> = < 0.0015, 0, 0></math>
'''Step 3'''
:Calculate the potential difference between B and A.
:<math>\Delta V_{BA} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{BA} = -(91179.912*0.0015 + 0*0 + 0*0) = -136.770V</math>
'''Step 4'''
:Calculate the displacement vector from A to D
:<math>\Delta l = < s_{1}, s_{2}, 0> - < s_{1}, 0, 0> = < 0, s_{2}, 0> = < 0, s_{2}, 0> = < 0, 0.006, 0></math>
'''Step 5'''
:Calculate the potential difference between A and D.
:<math>\Delta V_{AD} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AD} = -(91179.912*0 + 0*0 + 0*0) = 0V</math>
'''Step 6'''
:Add <math>\Delta V_{BA}</math> to <math>\Delta V_{AD}</math> to get <math>\Delta V_{BD}</math>
:<math>\Delta V_{BD} = \Delta V_{BA} + \Delta V_{AD} = -136.770 + 0 = -136.770V</math>
Notice that the voltage is negative since we are going from the positively charged plate (high potential) to the negatively charged plate (low potential) losing potential energy in the process.

Now you have to calculate the potential difference from A to C directly.

'''Step 1'''
:Calculate the displacement vector from B to D.
:<math>\Delta l = < s_{1}, s_{2}, 0> - < 0, 0, 0> = < s_{1}, s_{2}, 0> = < s_{1}, s_{2}, 0> = < 0.0015, 0.006, 0></math>
'''Step 2'''
:Calculate the potential difference between B and D.
:<math>\Delta V_{AD} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AD} = -(91179.912*0.0015 + 0*0 + 0*0) = -136.770V</math>
Notice that potential difference traveling from B to A then D is the same as the potential difference traveling from B to D directly. This exemplifies that the potential difference does not depend on the path taken to get from one point to another.

===Difficult===
[[File:Wiki_difficult.png|center]]
'''Problem''': A uniform spherical shell of charge +Q is centered at point B, as shown in the figure below. Show that ΔV = VC − VA is independent of path by calculating ΔV for each of these two paths (actually do the integrals). (Use the following as necessary: Q, R for the radius of the spherical shell, r1 for the radius of the circular arc A → D → C with B at its center, and ε0.)

'''Step 1'''
:Calculate the potential difference from A to C following the path A → B → C.
:First, calculate the potential difference from A to B.
:<math>\Delta V_{AB} = \int_{A}^{B} \vec{E} \cdot d\vec{l}</math>
:In order to calculate the potential difference from A to B, calculate the electric field of the sphere.
:<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}}</math>
:Now plug the electric field calculated above into the integral and solve.
:<math>\Delta V_{AB} = \int_{A}^{B} \vec{E} \cdot d\vec{r} = \int_{r_{1}}^{R} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{R^{2}} - \frac{Q}{r_{1}^{2}})</math>
:Now calculate the potential difference from B to C.
:Use the electric field of the sphere calculated above to solve the below integral.
:<math>\Delta V_{BC} = \int_{B}^{C} \vec{E} \cdot d\vec{l}</math>
:<math>\Delta V_{BC} = \int_{R}^{r_{1}} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{R^{2}})</math>
:Now add <math>\Delta V_{AB}</math> to <math>\Delta V_{BC}</math> to get <math>\Delta V_{AC}</math>
:<math>\Delta V_{AC} = \Delta V_{AB} + \Delta V_{BC} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{R^{2}} - \frac{Q}{r_{1}^{2}}) + \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{R^{2}}) = 0V </math>
'''Step 2'''
:Calculate the potential difference from A to C following the path A → D → C.
:First calculate the potential difference from A to D.
:<math>\Delta V_{AD} = \int_{A}^{D} \vec{E} \cdot d\vec{l}</math>
:Use the same electric field of the sphere calculated in step 1, shown below.
:<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}}</math>
:Now plug the electric field calculated above into the integral and solve.
:<math>\Delta V_{AD} = \int_{A}^{D} \vec{E} \cdot d\vec{r} = \int_{r_{1}}^{r_{1}} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}}) = 0V</math>
:Now calculate the potential difference from D to C.
:Use the electric field of the sphere calculated above to solve the below integral.
:<math>\Delta V_{DC} = \int_{D}^{C} \vec{E} \cdot d\vec{l}</math>
:<math>\Delta V_{DC} = \int_{r_{1}}^{r_{1}} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}})</math>
:Now add <math>\Delta V_{AD}</math> to <math>\Delta V_{DC}</math> to get <math>\Delta V_{AC}</math>
:<math>\Delta V_{AC} = \Delta V_{AD} + \Delta V_{DC} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}}) + \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}}) = 0V </math>
:Notice that either path taken from A to C results in the same potential difference of 0V.

==Connectedness==
#How is this topic connected to something that you are interested in?
::I am very interested in electronics and circuitry so knowing that potential difference is path independent definitely helps in determining the voltage between 2 points in a complicated circuit with many paths.
#How is it connected to your major?
::I am an electrical engineering major so this knowledge is very applicable to determining the voltage between 2 points in a circuit that contains many loops or is very complex.

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.
Luigi Galvani, an Italian physicist, discovered something he called "animal electricity" which was two different metals connected in series with a frog leg and each other. However, Alessandro Volta, another Italian physicist, realized that the frog leg was acting as a conductor and a detector of electricity, or an electrolyte. He later replaced the frog leg with brine soaked paper and noticed the same results. Through this, Volta discovered electromotive force relating to a galvanic cell, realizing that the difference between the two electrode potentials caused electricity to flow.

== See also ==

https://en.wikipedia.org/wiki/Voltage

===Further reading===

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elewor.html#c2

===External links===

http://mathwiki.ucdavis.edu/Core/Calculus/Vector_Calculus/Integration_in_Vector_Fields/Path_Independence,_Conservative_Fields,_and_Potential_Functions

==References==

Matter & Interactions volume II by Ruth W. Chabay and Bruce A. Sherwood.

[[Category:Energy]]

Potential Difference Path Independence, claimed by Aditya Mohile

2016-11-27T21:33:40Z

Zwright8:

''Claimed by Aditya Mohile''
'''Bold text'''
'''Claimed by Zachary Wright (Fall 2016)'''

Potential difference is a scalar quantity that measures the difference of electric potential between 2 points. The path taken from one point to another does not affect the potential difference therefore potential difference is said to be path independent.
==The Main Idea==

The potential difference between two points does not depend on the path taken to get from one point to another because the difference between the locations of the two points remain the same regardless of the path taken to get from one to another.

===A Mathematical Model===
:<math>\Delta V_{AB} = \int_{A}^{B} \vec{E} \cdot d\vec{l} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>

The potential difference, <math>\Delta V_{AB}</math>, is the integral of the electrical field from point A to point B. This integral, however, is the dot product of <math>\vec{E}</math> and <math>d\vec{l}</math> which is equal to the sum of the negative electric field multiplied by the difference of the x,y and z components of the 2 points, as see from the equation above. This proves that the path taken from point A to point B does not matter, only the difference in the positions of points A and B are important in calculating the potential difference.

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===
[[File:Wiki_simple.jpg|center]]
'''Problem''': Locations A, B, and C are in a region of uniform electric field, as shown in the diagram above. Location A is at < -0.6, 0, 0> m. Location B is at < 0.4, 0, 0>. Location C is at < 0.4, -0.3, 0>m. In the region the electric field = < -750, 400, 0> N/C. Calculate the potential difference between A and C traveling from A to B then to C. Calculate the potential difference traveling from A to C directly.

'''Step 1'''
:Calculate the displacement vector from A to B.
:<math>\Delta l = < 0.4, 0, 0> - < -0.6, 0, 0> = < 1, 0, 0></math>
'''Step 2'''
:Calculate the potential difference between A and B.
:<math>\Delta V_{AB} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AB} = -(-750*1 + 400*0 + 0*0) = 750V</math>
'''Step 3'''
:Calculate the displacement vector from B to C.
:<math>\Delta l = < 0.4, -0.3, 0> - < 0.4, 0, 0> = < 0, -0.3, 0></math>
'''Step 4'''
:Calculate the potential difference between B and C.
:<math>\Delta V_{BC} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{BC} = -(-750*0 + 400*-0.3 + 0*0) = 120V</math>
'''Step 5'''
:Add <math>\Delta V_{AB}</math> to <math>\Delta V_{BC}</math> to get <math>\Delta V_{AC}</math>
:<math>\Delta V_{AC} = \Delta V_{AB} + \Delta V_{BC} = 750 + 120 = 870V</math>
Now you have to calculate the potential difference from A to C directly.

'''Step 1'''
:Calculate the displacement vector from A to C.
:<math>\Delta l = < 0.4, -0.3, 0> - < -0.6, 0, 0> = < 1, -0.3, 0></math>
'''Step 2'''
:Calculate the potential difference between A and C.
:<math>\Delta V_{AB} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AB} = -(-750*1 + 400*-0.3 + 0*0) = 750 + 120 = 870V</math>
As you can see, potential difference from A to C directly is the same as the potential difference from A to B then to C.

===Middling===
[[File:Wiki_middling.jpg|350px|center]]
'''Problem''': A capacitor consists of two charged disks of radius 4.7 m separated by a distance s = 2 mm (see the figure). The magnitude of the charge on each disk is 56 µC. Consider points A, B, C, and D inside the capacitor, as shown in the diagram. The distance s1 = 1.5 mm, and the distance s2 = 0.6 mm. (Assume the +x axis is to the right, the +y axis is up, and the +z axis is out.) Find the potential difference between B and D, traveling through A. Find the potential difference from B to D directly.
'''Step 1'''
:Calculate the electric field inside the capacitor.
:<math>\vec E_{cap} = \frac{Q/A}{\epsilon_{0}}</math>
:<math>\vec E_{cap} = \frac{(56*10^{-6})/(\pi*4.7^{2})}{8.85*10^{-12}} = < 91179.912, 0, 0></math>
The electric field only has an x-component because the field travels from the positively charged plate to the negatively charged plate, traveling in the +x direction.

'''Step 2'''
:Calculate the displacement vector from B to A.
:Set the origin at B.
:<math>\Delta l = < s_{1}, 0, 0> - < 0, 0, 0> = < s_{1}, 0, 0> = < 0.0015, 0, 0></math>
'''Step 3'''
:Calculate the potential difference between B and A.
:<math>\Delta V_{BA} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{BA} = -(91179.912*0.0015 + 0*0 + 0*0) = -136.770V</math>
'''Step 4'''
:Calculate the displacement vector from A to D
:<math>\Delta l = < s_{1}, s_{2}, 0> - < s_{1}, 0, 0> = < 0, s_{2}, 0> = < 0, s_{2}, 0> = < 0, 0.006, 0></math>
'''Step 5'''
:Calculate the potential difference between A and D.
:<math>\Delta V_{AD} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AD} = -(91179.912*0 + 0*0 + 0*0) = 0V</math>
'''Step 6'''
:Add <math>\Delta V_{BA}</math> to <math>\Delta V_{AD}</math> to get <math>\Delta V_{BD}</math>
:<math>\Delta V_{BD} = \Delta V_{BA} + \Delta V_{AD} = -136.770 + 0 = -136.770V</math>
Notice that the voltage is negative since we are going from the positively charged plate (high potential) to the negatively charged plate (low potential) losing potential energy in the process.

Now you have to calculate the potential difference from A to C directly.

'''Step 1'''
:Calculate the displacement vector from B to D.
:<math>\Delta l = < s_{1}, s_{2}, 0> - < 0, 0, 0> = < s_{1}, s_{2}, 0> = < s_{1}, s_{2}, 0> = < 0.0015, 0.006, 0></math>
'''Step 2'''
:Calculate the potential difference between B and D.
:<math>\Delta V_{AD} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AD} = -(91179.912*0.0015 + 0*0 + 0*0) = -136.770V</math>
Notice that potential difference traveling from B to A then D is the same as the potential difference traveling from B to D directly. This exemplifies that the potential difference does not depend on the path taken to get from one point to another.

===Difficult===
[[File:Wiki_difficult.png|center]]
'''Problem''': A uniform spherical shell of charge +Q is centered at point B, as shown in the figure below. Show that ΔV = VC − VA is independent of path by calculating ΔV for each of these two paths (actually do the integrals). (Use the following as necessary: Q, R for the radius of the spherical shell, r1 for the radius of the circular arc A → D → C with B at its center, and ε0.)

'''Step 1'''
:Calculate the potential difference from A to C following the path A → B → C.
:First, calculate the potential difference from A to B.
:<math>\Delta V_{AB} = \int_{A}^{B} \vec{E} \cdot d\vec{l}</math>
:In order to calculate the potential difference from A to B, calculate the electric field of the sphere.
:<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}}</math>
:Now plug the electric field calculated above into the integral and solve.
:<math>\Delta V_{AB} = \int_{A}^{B} \vec{E} \cdot d\vec{r} = \int_{r_{1}}^{R} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{R^{2}} - \frac{Q}{r_{1}^{2}})</math>
:Now calculate the potential difference from B to C.
:Use the electric field of the sphere calculated above to solve the below integral.
:<math>\Delta V_{BC} = \int_{B}^{C} \vec{E} \cdot d\vec{l}</math>
:<math>\Delta V_{BC} = \int_{R}^{r_{1}} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{R^{2}})</math>
:Now add <math>\Delta V_{AB}</math> to <math>\Delta V_{BC}</math> to get <math>\Delta V_{AC}</math>
:<math>\Delta V_{AC} = \Delta V_{AB} + \Delta V_{BC} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{R^{2}} - \frac{Q}{r_{1}^{2}}) + \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{R^{2}}) = 0V </math>
'''Step 2'''
:Calculate the potential difference from A to C following the path A → D → C.
:First calculate the potential difference from A to D.
:<math>\Delta V_{AD} = \int_{A}^{D} \vec{E} \cdot d\vec{l}</math>
:Use the same electric field of the sphere calculated in step 1, shown below.
:<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}}</math>
:Now plug the electric field calculated above into the integral and solve.
:<math>\Delta V_{AD} = \int_{A}^{D} \vec{E} \cdot d\vec{r} = \int_{r_{1}}^{r_{1}} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}}) = 0V</math>
:Now calculate the potential difference from D to C.
:Use the electric field of the sphere calculated above to solve the below integral.
:<math>\Delta V_{DC} = \int_{D}^{C} \vec{E} \cdot d\vec{l}</math>
:<math>\Delta V_{DC} = \int_{r_{1}}^{r_{1}} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}})</math>
:Now add <math>\Delta V_{AD}</math> to <math>\Delta V_{DC}</math> to get <math>\Delta V_{AC}</math>
:<math>\Delta V_{AC} = \Delta V_{AD} + \Delta V_{DC} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}}) + \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}}) = 0V </math>
:Notice that either path taken from A to C results in the same potential difference of 0V.

==Connectedness==
#How is this topic connected to something that you are interested in?
::I am very interested in electronics and circuitry so knowing that potential difference is path independent definitely helps in determining the voltage between 2 points in a complicated circuit with many paths.
#How is it connected to your major?
::I am an electrical engineering major so this knowledge is very applicable to determining the voltage between 2 points in a circuit that contains many loops or is very complex.

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.
Luigi Galvani, an Italian physicist, discovered something he called "animal electricity" which was two different metals connected in series with a frog leg and each other. However, Alessandro Volta, another Italian physicist, realized that the frog leg was acting as a conductor and a detector of electricity, or an electrolyte. He later replaced the frog leg with brine soaked paper and noticed the same results. Through this, Volta discovered electromotive force relating to a galvanic cell, realizing that the difference between the two electrode potentials caused electricity to flow.

== See also ==

https://en.wikipedia.org/wiki/Voltage

===Further reading===

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elewor.html#c2

===External links===

http://mathwiki.ucdavis.edu/Core/Calculus/Vector_Calculus/Integration_in_Vector_Fields/Path_Independence,_Conservative_Fields,_and_Potential_Functions

==References==

Matter & Interactions volume II by Ruth W. Chabay and Bruce A. Sherwood.

[[Category:Energy]]

Potential Difference Path Independence, claimed by Aditya Mohile

2016-11-27T21:33:26Z

Zwright8:

''Claimed by Aditya Mohile''
'''Claimed by Zachary Wright (Fall 2016)'''

Potential difference is a scalar quantity that measures the difference of electric potential between 2 points. The path taken from one point to another does not affect the potential difference therefore potential difference is said to be path independent.
==The Main Idea==

The potential difference between two points does not depend on the path taken to get from one point to another because the difference between the locations of the two points remain the same regardless of the path taken to get from one to another.

===A Mathematical Model===
:<math>\Delta V_{AB} = \int_{A}^{B} \vec{E} \cdot d\vec{l} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>

The potential difference, <math>\Delta V_{AB}</math>, is the integral of the electrical field from point A to point B. This integral, however, is the dot product of <math>\vec{E}</math> and <math>d\vec{l}</math> which is equal to the sum of the negative electric field multiplied by the difference of the x,y and z components of the 2 points, as see from the equation above. This proves that the path taken from point A to point B does not matter, only the difference in the positions of points A and B are important in calculating the potential difference.

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===
[[File:Wiki_simple.jpg|center]]
'''Problem''': Locations A, B, and C are in a region of uniform electric field, as shown in the diagram above. Location A is at < -0.6, 0, 0> m. Location B is at < 0.4, 0, 0>. Location C is at < 0.4, -0.3, 0>m. In the region the electric field = < -750, 400, 0> N/C. Calculate the potential difference between A and C traveling from A to B then to C. Calculate the potential difference traveling from A to C directly.

'''Step 1'''
:Calculate the displacement vector from A to B.
:<math>\Delta l = < 0.4, 0, 0> - < -0.6, 0, 0> = < 1, 0, 0></math>
'''Step 2'''
:Calculate the potential difference between A and B.
:<math>\Delta V_{AB} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AB} = -(-750*1 + 400*0 + 0*0) = 750V</math>
'''Step 3'''
:Calculate the displacement vector from B to C.
:<math>\Delta l = < 0.4, -0.3, 0> - < 0.4, 0, 0> = < 0, -0.3, 0></math>
'''Step 4'''
:Calculate the potential difference between B and C.
:<math>\Delta V_{BC} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{BC} = -(-750*0 + 400*-0.3 + 0*0) = 120V</math>
'''Step 5'''
:Add <math>\Delta V_{AB}</math> to <math>\Delta V_{BC}</math> to get <math>\Delta V_{AC}</math>
:<math>\Delta V_{AC} = \Delta V_{AB} + \Delta V_{BC} = 750 + 120 = 870V</math>
Now you have to calculate the potential difference from A to C directly.

'''Step 1'''
:Calculate the displacement vector from A to C.
:<math>\Delta l = < 0.4, -0.3, 0> - < -0.6, 0, 0> = < 1, -0.3, 0></math>
'''Step 2'''
:Calculate the potential difference between A and C.
:<math>\Delta V_{AB} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AB} = -(-750*1 + 400*-0.3 + 0*0) = 750 + 120 = 870V</math>
As you can see, potential difference from A to C directly is the same as the potential difference from A to B then to C.

===Middling===
[[File:Wiki_middling.jpg|350px|center]]
'''Problem''': A capacitor consists of two charged disks of radius 4.7 m separated by a distance s = 2 mm (see the figure). The magnitude of the charge on each disk is 56 µC. Consider points A, B, C, and D inside the capacitor, as shown in the diagram. The distance s1 = 1.5 mm, and the distance s2 = 0.6 mm. (Assume the +x axis is to the right, the +y axis is up, and the +z axis is out.) Find the potential difference between B and D, traveling through A. Find the potential difference from B to D directly.
'''Step 1'''
:Calculate the electric field inside the capacitor.
:<math>\vec E_{cap} = \frac{Q/A}{\epsilon_{0}}</math>
:<math>\vec E_{cap} = \frac{(56*10^{-6})/(\pi*4.7^{2})}{8.85*10^{-12}} = < 91179.912, 0, 0></math>
The electric field only has an x-component because the field travels from the positively charged plate to the negatively charged plate, traveling in the +x direction.

'''Step 2'''
:Calculate the displacement vector from B to A.
:Set the origin at B.
:<math>\Delta l = < s_{1}, 0, 0> - < 0, 0, 0> = < s_{1}, 0, 0> = < 0.0015, 0, 0></math>
'''Step 3'''
:Calculate the potential difference between B and A.
:<math>\Delta V_{BA} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{BA} = -(91179.912*0.0015 + 0*0 + 0*0) = -136.770V</math>
'''Step 4'''
:Calculate the displacement vector from A to D
:<math>\Delta l = < s_{1}, s_{2}, 0> - < s_{1}, 0, 0> = < 0, s_{2}, 0> = < 0, s_{2}, 0> = < 0, 0.006, 0></math>
'''Step 5'''
:Calculate the potential difference between A and D.
:<math>\Delta V_{AD} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AD} = -(91179.912*0 + 0*0 + 0*0) = 0V</math>
'''Step 6'''
:Add <math>\Delta V_{BA}</math> to <math>\Delta V_{AD}</math> to get <math>\Delta V_{BD}</math>
:<math>\Delta V_{BD} = \Delta V_{BA} + \Delta V_{AD} = -136.770 + 0 = -136.770V</math>
Notice that the voltage is negative since we are going from the positively charged plate (high potential) to the negatively charged plate (low potential) losing potential energy in the process.

Now you have to calculate the potential difference from A to C directly.

'''Step 1'''
:Calculate the displacement vector from B to D.
:<math>\Delta l = < s_{1}, s_{2}, 0> - < 0, 0, 0> = < s_{1}, s_{2}, 0> = < s_{1}, s_{2}, 0> = < 0.0015, 0.006, 0></math>
'''Step 2'''
:Calculate the potential difference between B and D.
:<math>\Delta V_{AD} = -(E_{x}\Delta x + E_{y}\Delta y + E_{z}\Delta z)</math>
:<math>\Delta V_{AD} = -(91179.912*0.0015 + 0*0 + 0*0) = -136.770V</math>
Notice that potential difference traveling from B to A then D is the same as the potential difference traveling from B to D directly. This exemplifies that the potential difference does not depend on the path taken to get from one point to another.

===Difficult===
[[File:Wiki_difficult.png|center]]
'''Problem''': A uniform spherical shell of charge +Q is centered at point B, as shown in the figure below. Show that ΔV = VC − VA is independent of path by calculating ΔV for each of these two paths (actually do the integrals). (Use the following as necessary: Q, R for the radius of the spherical shell, r1 for the radius of the circular arc A → D → C with B at its center, and ε0.)

'''Step 1'''
:Calculate the potential difference from A to C following the path A → B → C.
:First, calculate the potential difference from A to B.
:<math>\Delta V_{AB} = \int_{A}^{B} \vec{E} \cdot d\vec{l}</math>
:In order to calculate the potential difference from A to B, calculate the electric field of the sphere.
:<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}}</math>
:Now plug the electric field calculated above into the integral and solve.
:<math>\Delta V_{AB} = \int_{A}^{B} \vec{E} \cdot d\vec{r} = \int_{r_{1}}^{R} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{R^{2}} - \frac{Q}{r_{1}^{2}})</math>
:Now calculate the potential difference from B to C.
:Use the electric field of the sphere calculated above to solve the below integral.
:<math>\Delta V_{BC} = \int_{B}^{C} \vec{E} \cdot d\vec{l}</math>
:<math>\Delta V_{BC} = \int_{R}^{r_{1}} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{R^{2}})</math>
:Now add <math>\Delta V_{AB}</math> to <math>\Delta V_{BC}</math> to get <math>\Delta V_{AC}</math>
:<math>\Delta V_{AC} = \Delta V_{AB} + \Delta V_{BC} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{R^{2}} - \frac{Q}{r_{1}^{2}}) + \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{R^{2}}) = 0V </math>
'''Step 2'''
:Calculate the potential difference from A to C following the path A → D → C.
:First calculate the potential difference from A to D.
:<math>\Delta V_{AD} = \int_{A}^{D} \vec{E} \cdot d\vec{l}</math>
:Use the same electric field of the sphere calculated in step 1, shown below.
:<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}}</math>
:Now plug the electric field calculated above into the integral and solve.
:<math>\Delta V_{AD} = \int_{A}^{D} \vec{E} \cdot d\vec{r} = \int_{r_{1}}^{r_{1}} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}}) = 0V</math>
:Now calculate the potential difference from D to C.
:Use the electric field of the sphere calculated above to solve the below integral.
:<math>\Delta V_{DC} = \int_{D}^{C} \vec{E} \cdot d\vec{l}</math>
:<math>\Delta V_{DC} = \int_{r_{1}}^{r_{1}} \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}} \cdot d\vec{r} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}})</math>
:Now add <math>\Delta V_{AD}</math> to <math>\Delta V_{DC}</math> to get <math>\Delta V_{AC}</math>
:<math>\Delta V_{AC} = \Delta V_{AD} + \Delta V_{DC} = \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}}) + \frac{1}{4\pi\epsilon_{0}}(\frac{Q}{r_{1}^{2}} - \frac{Q}{r_{1}^{2}}) = 0V </math>
:Notice that either path taken from A to C results in the same potential difference of 0V.

==Connectedness==
#How is this topic connected to something that you are interested in?
::I am very interested in electronics and circuitry so knowing that potential difference is path independent definitely helps in determining the voltage between 2 points in a complicated circuit with many paths.
#How is it connected to your major?
::I am an electrical engineering major so this knowledge is very applicable to determining the voltage between 2 points in a circuit that contains many loops or is very complex.

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.
Luigi Galvani, an Italian physicist, discovered something he called "animal electricity" which was two different metals connected in series with a frog leg and each other. However, Alessandro Volta, another Italian physicist, realized that the frog leg was acting as a conductor and a detector of electricity, or an electrolyte. He later replaced the frog leg with brine soaked paper and noticed the same results. Through this, Volta discovered electromotive force relating to a galvanic cell, realizing that the difference between the two electrode potentials caused electricity to flow.

== See also ==

https://en.wikipedia.org/wiki/Voltage

===Further reading===

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elewor.html#c2

===External links===

http://mathwiki.ucdavis.edu/Core/Calculus/Vector_Calculus/Integration_in_Vector_Fields/Path_Independence,_Conservative_Fields,_and_Potential_Functions

==References==

Matter & Interactions volume II by Ruth W. Chabay and Bruce A. Sherwood.

[[Category:Energy]]

Electric Potential

2016-11-27T21:31:36Z

Zwright8:

'''AUTHOR: Rmohammed7'''

'''REVISED BY: HAYOUNG KIM (SPRING 2016)'''

This page discusses the Electric Potential and examples of how it is used.

----

== The Main Idea ==

Electric potential is a rather difficult concept as it is usually accompanied by other topics. For example, electric potential is not the same as electric potential energy as electric potential is defined as the total potential energy per charge and is basically used to describe the electric field's effect at a certain location. In other words, electric potential is purely dependent on the electric field (whether it is uniform or nonuniform) and the location, whereas the electric potential energy also depends on the amount of charge the object in the system is experiencing. Also, although electric potential is an important topic to learn, most problems encountered will not ask to find just the "electric potential," instead, questions will most likely ask for the "electric potential difference." This is because electric potential is measured using different locations, or more specifically pathways between the different locations, so instead of determining the electric potential of location A and the electric potential of final location B, it would make more sense to determine the "difference in electric potential between locations A and B."

===A Mathematical Model===

Like mentioned in the Main Idea, instead of electric potential, in most cases, electric potential difference is needed to be found. The general equation for the potential difference is '''<math>∆{{U}_{electric}} = {q} * ∆{V} </math>'''.

'''<math>∆{{U}_{electric}}</math>''' is the electric potential energy, which is measured in Joules (J). '''q''' is the charge of the particle moving through the path of the electric potential difference, which is measured in coulombs (C). '''∆V''' is the electric potential difference, which is measured in Joules per Coulomb (J/C), or just Volts (V).

Aside from the general equation, the electric potential difference can also be found in other ways. The potential difference in an '''uniform field''' is '''<math>∆{V} = -({E}_{x}∆{x} + {E}_{y}∆{y} + {E}_{z}∆{z})</math>''', which can also be written as '''<math>∆{V} = -\vec{E}·∆\vec{l}</math>'''.

'''∆V''' is the electric potential difference, which is measured in Joules per Coulomb (J/C), or just Volts (V). '''E''' is the electric field, which is measured in Newtons per Coulomb (N/C), and it is important to note that the different direction components of the electric field are used in the equation. '''l''' (or the '''x''', '''y''', '''z''') is the distance between the two described locations, which is measured in meters, and x, y, and z, are the different components of the difference.

The electric potential difference in a '''nonuniform field''' is '''<math>∆{V} = -∑ \vec{E}·∆\vec{l}</math>'''. The different parts in this particular equation resembles the equation for the potential difference in an uniform field, except that with the nonuniform field, the potential difference in the different fields are summed up. This situation can be quite easy, but when the system gets difficult, first, choose a path and divide it into smaller pieces of '''<math>∆\vec{l}</math>'''; second, write an expression for '''<math>∆{V} = -\vec{E}·∆\vec{l}</math>''' of one piece; third, add up the contributions of all the pieces; last, check the result to make sure the magnitude, direction, and units make sense.

Aside from just calculating the value of the electric potential difference, determining the sign is also quite crucial to be successful. If the path being considered is in the same direction as the electric field, then the sign with be negative (-), or the potential is decreasing. If the path being considered is in the opposite direction as the electric field, then the sign will be positive (+), or the potential is increasing. If the path being considered is perpendicular to the electric field, then the potential difference will just be zero and have no direction. With these simple tips, the direction of the potential difference can be rechecked with the answer calculated using vectors.

Also, when working with different situations, it is nice to keep in mind that in a conductor, the electric field is zero. Therefore, the potential difference is zero as well. In an insulator, the electric field is '''<math>{E}_{applied} / K</math>''' where '''K''' is the dielectric constant. Also, the round trip potential difference is always zero, or in other words, if you start from a certain point and end at the same point, then, the potential difference will be zero.

===A Computational Model===
Click on the link to see Electric Potential through VPython!

Make sure to press "Run" to see the principle in action!

[https://trinket.io/embed/glowscript/0a7e486c94 Teach hand-on with GlowScript]

Watch this video for a more visual approach!

[https://www.youtube.com/watch?v=-Rb9guSEeVE Electric Potential: Visualizing Voltage with 3D animations]

==Examples==

===Simple===
In a capacitor, the negative charges are located on the left plate, and the positive charges are located on the right plate. Location A is at the left end of the capacitor, and Location B is at the right end of the capacitor, or in other words, Location A and B are only different in terms of their x component location. The path moves from A to B. What is the direction of the electric field? Is the potential difference positive or negative? [Hint: Draw a picture!]

'''Answer: The electric field is to the left. The potential difference is increasing, or is positive.'''

'''Explanation:'''
The electric field always moves away from the positive charge and towards the negative charge, which means the electric field in this example is to the left. Because the direction and the electric field and the direction of the path are opposite, the potential difference is increasing, or is positive.

===Middling===

Calculate the difference in electric potential between two locations A, which is at <-0.4, 0,0>m, and B, which is at <0.2,0,0>m. The electric field in the location is <500,0,0> N/C.

'''Answer: -300V'''

'''Explanation:'''

<math>∆\vec{l}</math> = <0.2,0,0>m - <-0.4,0,0>m = <0.6,0,0>m

<math>∆{V} = -({E}_{x}∆{x} + {E}_{y}∆{y} + {E}_{z}∆{z})</math>

<math>∆{V} = -(500 N/C * 0.6 m + 0*0 + 0*0)</math>

<math>∆{V} = -300 V</math>

===Difficult===

Suppose that from x=0 to x=3 the electric field is uniform and given by

'''Answer: -14.3 V ; 2.3e-18 J'''

'''Explanation:'''

<math>∆{V} = -\int_C^D {E}_{x} \, dx </math>

<math>∆{V} = -\int_{1e-10}^{2e-8} \tfrac{1}{4π{ε}_{0}}*\tfrac{Q}{{x}^{2}} \, dx </math>

<math>∆{V} = \tfrac{1}{4π{ε}_{0}}*{1.6e-19 C}*({\tfrac{1}{2e-8 m} - \tfrac{1}{1e-10 m}}) </math>

<math>∆{V}</math> = '''-14.3 V'''

∆K + ∆U = <math>{W}_{ext}</math>

<math>{W}_{ext}</math> = 0 + (-e)(∆V)

<math>{W}_{ext}</math> = (-1.6e-19 C)(-14.3 V)

<math>{W}_{ext}</math> = '''2.3e-18 J'''

==Connectedness==
'''How is this topic connected to something that you are interested in?'''

[A] I am interested in robotic systems and building circuit boards and electrical systems for manufacturing robots. While studying this section in the book, I was able to connect back many of the concepts and calculations back to robotics and the electrical component of automated systems.

[R] Since high school, I never really understood how to work with the voltmeter and what it measured, and I have always wanted to know, but although this particular wiki page did not go into the details and other branches of electric potential, it led me to find the answers to something I was interested in since high school, the concept of electric potential.

'''How is it connected to your major?'''

[A] I am a Mechanical Engineering major, so I will be dealing with the electrical components of machines when I work. Therefore, I have to know these certain concepts such as electric potential in order to fully understand how they work and interact.

[R] As a biochemistry major, electric potential and electric potential difference is not particularly related to my major, but in chemistry classes, we use electrostatic potential maps (electrostatic potential energy maps) that shows the charge distributions throughout a molecule. Although the main use in electric potential is different in physics and biochemistry (where physicists use it identify the effect of the electric field at a location), I still found it interesting as the concept of electric potential (buildup) was being used in quite a different way.

'''Is there an interesting industrial application?'''

[A] Electrical potential is used to find the voltage across a path. This is useful when working with circuit components and attempting to manipulate the power output or current throughout a component.

[R] Electric potential sensors are being used to detect a variety of electrical signals made by the human body, thus contributing to the field of electrophysiology.

==History==

The idea of electric potential, in a way, started with Ben Franklin and his experiments in the 1740s. He began to understand the flow of electricity, which eventually paved the path towards explaining electric potential and potential difference. Scientists finally began to understand how electric fields were actually affecting the charges and the surrounding environment. Benjamin Franklin first shocked himself in 1746, while conducting experiments on electricity with found objects from around his house. Six years later, or 261 years ago for us, the founding father flew a kite attached to a key and a silk ribbon in a thunderstorm and effectively trapped lightning in a jar. The experiment is now seen as a watershed moment in mankind's venture to channel a force of nature that was viewed quite abstractly.

By the time Franklin started experimenting with electricity, he'd already found fame and fortune as the author of Poor Richard's Almanack. Electricity wasn't a very well understood phenomenon at that point, so Franklin's research proved to be fairly foundational. The early experiments, experts believe, were inspired by other scientists' work and the shortcomings therein.

That early brush with the dangers of electricity left an impression on Franklin. He described the sensation as "a universal blow throughout my whole body from head to foot, which seemed within as well as without; after which the first thing I took notice of was a violent quick shaking of my body." However, it didn't scare him away. In the handful of years before his famous kite experiment, Franklin contributed everything from designing the first battery designs to establishing some common nomenclature in the study of electricity. Although Franklin is often coined the father of electricity, after he set the foundations of electricity, many other scientists contributed his or her research in the advancement of electricity and eventually led to the discovery of electric potential and potential difference.

== See also ==

Like mentioned multiple times throughout the page, although electric potential is a huge and important topic, it has many branches, which makes the concept of electric potential difficult to stand alone. Even with this page, to support the concept of electric potential, many crucial branches of the topic appeared, like potential difference (which also branched into [[http://www.physicsbook.gatech.edu/Potential_Difference_Path_Independence,_claimed_by_Aditya_Mohile Potential Difference Path Independence]], [[http://www.physicsbook.gatech.edu/Potential_Difference_in_a_Uniform_Field Potential Difference In A Uniform Field]], and [[http://www.physicsbook.gatech.edu/Potential_Difference_of_Point_Charge_in_a_Non-Uniform_Field Potential Difference In A Nonuniform Field]]).

===External Links===

[1] https://www.khanacademy.org/test-prep/mcat/physical-processes/electrostatics-1/v/electric-potential-at-a-point-in-space

[2] https://www.youtube.com/watch?v=pcWz4tP_zUw

[3] https://www.youtube.com/watch?v=Vpa_uApmNoo

==References==

[1] "Benjamin Franklin and Electricity." Benjamin Franklin and Electricity. N.p., n.d. Web. 17 Apr. 2016. <http://www.americaslibrary.gov/aa/franklinb/aa_franklinb_electric_1.html>.

[2] Bottyan, Thomas. "Electrostatic Potential Maps." Chemwiki. N.p., 02 Oct. 2013. Web. 17 Apr. 2016. <http://chemwiki.ucdavis.edu/Core/Theoretical_Chemistry/Chemical_Bonding/General_Principles_of_Chemical_Bonding/Electrostatic_Potential_maps>.

[3] "Electric Potential Difference." Electric Potential Difference. The Physics Classroom, n.d. Web. 14 Apr. 2016. <http://www.physicsclassroom.com/class/circuits/Lesson-1/Electric-Potential-Difference>.

[4] Harland, C. J., T. D. Clark, and R. J. Prance. "Applications of Electric Potential (Displacement Current) Sensors in Human Body Electrophysiology." International Society for Industrial Process Tomography, n.d. Web. 16 Apr. 2016. <http://www.isipt.org/world-congress/3/269.html>.

[5] Sherwood, Bruce A. "2.1 The Momentum Principle." Matter & Interactions. By Ruth W. Chabay. 4th ed. Vol. 1. N.p.: John Wiley & Sons, 2015. 45-50. Print. Modern Mechanics.

Electric Potential

2016-11-27T21:27:33Z

Zwright8:

'''AUTHOR: Rmohammed7'''

'''REVISED BY: HAYOUNG KIM (SPRING 2016)'''

'''REVISED BY: ZACHARY WRIGHT ( FALL 2016)'''

This page discusses the Electric Potential and examples of how it is used.

----

== The Main Idea ==

Electric potential is a rather difficult concept as it is usually accompanied by other topics. For example, electric potential is not the same as electric potential energy as electric potential is defined as the total potential energy per charge and is basically used to describe the electric field's effect at a certain location. In other words, electric potential is purely dependent on the electric field (whether it is uniform or nonuniform) and the location, whereas the electric potential energy also depends on the amount of charge the object in the system is experiencing. Also, although electric potential is an important topic to learn, most problems encountered will not ask to find just the "electric potential," instead, questions will most likely ask for the "electric potential difference." This is because electric potential is measured using different locations, or more specifically pathways between the different locations, so instead of determining the electric potential of location A and the electric potential of final location B, it would make more sense to determine the "difference in electric potential between locations A and B."

===A Mathematical Model===

Like mentioned in the Main Idea, instead of electric potential, in most cases, electric potential difference is needed to be found. The general equation for the potential difference is '''<math>∆{{U}_{electric}} = {q} * ∆{V} </math>'''.

'''<math>∆{{U}_{electric}}</math>''' is the electric potential energy, which is measured in Joules (J). '''q''' is the charge of the particle moving through the path of the electric potential difference, which is measured in coulombs (C). '''∆V''' is the electric potential difference, which is measured in Joules per Coulomb (J/C), or just Volts (V).

Aside from the general equation, the electric potential difference can also be found in other ways. The potential difference in an '''uniform field''' is '''<math>∆{V} = -({E}_{x}∆{x} + {E}_{y}∆{y} + {E}_{z}∆{z})</math>''', which can also be written as '''<math>∆{V} = -\vec{E}·∆\vec{l}</math>'''.

'''∆V''' is the electric potential difference, which is measured in Joules per Coulomb (J/C), or just Volts (V). '''E''' is the electric field, which is measured in Newtons per Coulomb (N/C), and it is important to note that the different direction components of the electric field are used in the equation. '''l''' (or the '''x''', '''y''', '''z''') is the distance between the two described locations, which is measured in meters, and x, y, and z, are the different components of the difference.

The electric potential difference in a '''nonuniform field''' is '''<math>∆{V} = -∑ \vec{E}·∆\vec{l}</math>'''. The different parts in this particular equation resembles the equation for the potential difference in an uniform field, except that with the nonuniform field, the potential difference in the different fields are summed up. This situation can be quite easy, but when the system gets difficult, first, choose a path and divide it into smaller pieces of '''<math>∆\vec{l}</math>'''; second, write an expression for '''<math>∆{V} = -\vec{E}·∆\vec{l}</math>''' of one piece; third, add up the contributions of all the pieces; last, check the result to make sure the magnitude, direction, and units make sense.

Aside from just calculating the value of the electric potential difference, determining the sign is also quite crucial to be successful. If the path being considered is in the same direction as the electric field, then the sign with be negative (-), or the potential is decreasing. If the path being considered is in the opposite direction as the electric field, then the sign will be positive (+), or the potential is increasing. If the path being considered is perpendicular to the electric field, then the potential difference will just be zero and have no direction. With these simple tips, the direction of the potential difference can be rechecked with the answer calculated using vectors.

Also, when working with different situations, it is nice to keep in mind that in a conductor, the electric field is zero. Therefore, the potential difference is zero as well. In an insulator, the electric field is '''<math>{E}_{applied} / K</math>''' where '''K''' is the dielectric constant. Also, the round trip potential difference is always zero, or in other words, if you start from a certain point and end at the same point, then, the potential difference will be zero.

===A Computational Model===
Click on the link to see Electric Potential through VPython!

Make sure to press "Run" to see the principle in action!

[https://trinket.io/embed/glowscript/0a7e486c94 Teach hand-on with GlowScript]

Watch this video for a more visual approach!

[https://www.youtube.com/watch?v=-Rb9guSEeVE Electric Potential: Visualizing Voltage with 3D animations]

==Examples==

===Simple===
In a capacitor, the negative charges are located on the left plate, and the positive charges are located on the right plate. Location A is at the left end of the capacitor, and Location B is at the right end of the capacitor, or in other words, Location A and B are only different in terms of their x component location. The path moves from A to B. What is the direction of the electric field? Is the potential difference positive or negative? [Hint: Draw a picture!]

'''Answer: The electric field is to the left. The potential difference is increasing, or is positive.'''

'''Explanation:'''
The electric field always moves away from the positive charge and towards the negative charge, which means the electric field in this example is to the left. Because the direction and the electric field and the direction of the path are opposite, the potential difference is increasing, or is positive.

===Middling===

Calculate the difference in electric potential between two locations A, which is at <-0.4, 0,0>m, and B, which is at <0.2,0,0>m. The electric field in the location is <500,0,0> N/C.

'''Answer: -300V'''

'''Explanation:'''

<math>∆\vec{l}</math> = <0.2,0,0>m - <-0.4,0,0>m = <0.6,0,0>m

<math>∆{V} = -({E}_{x}∆{x} + {E}_{y}∆{y} + {E}_{z}∆{z})</math>

<math>∆{V} = -(500 N/C * 0.6 m + 0*0 + 0*0)</math>

<math>∆{V} = -300 V</math>

===Difficult===

Suppose that from x=0 to x=3 the electric field is uniform and given by

'''Answer: -14.3 V ; 2.3e-18 J'''

'''Explanation:'''

<math>∆{V} = -\int_C^D {E}_{x} \, dx </math>

<math>∆{V} = -\int_{1e-10}^{2e-8} \tfrac{1}{4π{ε}_{0}}*\tfrac{Q}{{x}^{2}} \, dx </math>

<math>∆{V} = \tfrac{1}{4π{ε}_{0}}*{1.6e-19 C}*({\tfrac{1}{2e-8 m} - \tfrac{1}{1e-10 m}}) </math>

<math>∆{V}</math> = '''-14.3 V'''

∆K + ∆U = <math>{W}_{ext}</math>

<math>{W}_{ext}</math> = 0 + (-e)(∆V)

<math>{W}_{ext}</math> = (-1.6e-19 C)(-14.3 V)

<math>{W}_{ext}</math> = '''2.3e-18 J'''

==Connectedness==
'''How is this topic connected to something that you are interested in?'''

[A] I am interested in robotic systems and building circuit boards and electrical systems for manufacturing robots. While studying this section in the book, I was able to connect back many of the concepts and calculations back to robotics and the electrical component of automated systems.

[R] Since high school, I never really understood how to work with the voltmeter and what it measured, and I have always wanted to know, but although this particular wiki page did not go into the details and other branches of electric potential, it led me to find the answers to something I was interested in since high school, the concept of electric potential.

'''How is it connected to your major?'''

[A] I am a Mechanical Engineering major, so I will be dealing with the electrical components of machines when I work. Therefore, I have to know these certain concepts such as electric potential in order to fully understand how they work and interact.

[R] As a biochemistry major, electric potential and electric potential difference is not particularly related to my major, but in chemistry classes, we use electrostatic potential maps (electrostatic potential energy maps) that shows the charge distributions throughout a molecule. Although the main use in electric potential is different in physics and biochemistry (where physicists use it identify the effect of the electric field at a location), I still found it interesting as the concept of electric potential (buildup) was being used in quite a different way.

'''Is there an interesting industrial application?'''

[A] Electrical potential is used to find the voltage across a path. This is useful when working with circuit components and attempting to manipulate the power output or current throughout a component.

[R] Electric potential sensors are being used to detect a variety of electrical signals made by the human body, thus contributing to the field of electrophysiology.

==History==

The idea of electric potential, in a way, started with Ben Franklin and his experiments in the 1740s. He began to understand the flow of electricity, which eventually paved the path towards explaining electric potential and potential difference. Scientists finally began to understand how electric fields were actually affecting the charges and the surrounding environment. Benjamin Franklin first shocked himself in 1746, while conducting experiments on electricity with found objects from around his house. Six years later, or 261 years ago for us, the founding father flew a kite attached to a key and a silk ribbon in a thunderstorm and effectively trapped lightning in a jar. The experiment is now seen as a watershed moment in mankind's venture to channel a force of nature that was viewed quite abstractly.

By the time Franklin started experimenting with electricity, he'd already found fame and fortune as the author of Poor Richard's Almanack. Electricity wasn't a very well understood phenomenon at that point, so Franklin's research proved to be fairly foundational. The early experiments, experts believe, were inspired by other scientists' work and the shortcomings therein.

That early brush with the dangers of electricity left an impression on Franklin. He described the sensation as "a universal blow throughout my whole body from head to foot, which seemed within as well as without; after which the first thing I took notice of was a violent quick shaking of my body." However, it didn't scare him away. In the handful of years before his famous kite experiment, Franklin contributed everything from designing the first battery designs to establishing some common nomenclature in the study of electricity. Although Franklin is often coined the father of electricity, after he set the foundations of electricity, many other scientists contributed his or her research in the advancement of electricity and eventually led to the discovery of electric potential and potential difference.

== See also ==

Like mentioned multiple times throughout the page, although electric potential is a huge and important topic, it has many branches, which makes the concept of electric potential difficult to stand alone. Even with this page, to support the concept of electric potential, many crucial branches of the topic appeared, like potential difference (which also branched into [[http://www.physicsbook.gatech.edu/Potential_Difference_Path_Independence,_claimed_by_Aditya_Mohile Potential Difference Path Independence]], [[http://www.physicsbook.gatech.edu/Potential_Difference_in_a_Uniform_Field Potential Difference In A Uniform Field]], and [[http://www.physicsbook.gatech.edu/Potential_Difference_of_Point_Charge_in_a_Non-Uniform_Field Potential Difference In A Nonuniform Field]]).

===External Links===

[1] https://www.khanacademy.org/test-prep/mcat/physical-processes/electrostatics-1/v/electric-potential-at-a-point-in-space

[2] https://www.youtube.com/watch?v=pcWz4tP_zUw

[3] https://www.youtube.com/watch?v=Vpa_uApmNoo

==References==

[1] "Benjamin Franklin and Electricity." Benjamin Franklin and Electricity. N.p., n.d. Web. 17 Apr. 2016. <http://www.americaslibrary.gov/aa/franklinb/aa_franklinb_electric_1.html>.

[2] Bottyan, Thomas. "Electrostatic Potential Maps." Chemwiki. N.p., 02 Oct. 2013. Web. 17 Apr. 2016. <http://chemwiki.ucdavis.edu/Core/Theoretical_Chemistry/Chemical_Bonding/General_Principles_of_Chemical_Bonding/Electrostatic_Potential_maps>.

[3] "Electric Potential Difference." Electric Potential Difference. The Physics Classroom, n.d. Web. 14 Apr. 2016. <http://www.physicsclassroom.com/class/circuits/Lesson-1/Electric-Potential-Difference>.

[4] Harland, C. J., T. D. Clark, and R. J. Prance. "Applications of Electric Potential (Displacement Current) Sensors in Human Body Electrophysiology." International Society for Industrial Process Tomography, n.d. Web. 16 Apr. 2016. <http://www.isipt.org/world-congress/3/269.html>.

[5] Sherwood, Bruce A. "2.1 The Momentum Principle." Matter & Interactions. By Ruth W. Chabay. 4th ed. Vol. 1. N.p.: John Wiley & Sons, 2015. 45-50. Print. Modern Mechanics.

Electric Potential

2016-11-27T21:26:52Z

Zwright8:

'''AUTHOR: Rmohammed7'''

'''REVISED BY: HAYOUNG KIM (SPRING 2016)'''

""REVISED BY: ZACHARY WRIGHT ( FALL 2016)""

This page discusses the Electric Potential and examples of how it is used.

----

== The Main Idea ==

Electric potential is a rather difficult concept as it is usually accompanied by other topics. For example, electric potential is not the same as electric potential energy as electric potential is defined as the total potential energy per charge and is basically used to describe the electric field's effect at a certain location. In other words, electric potential is purely dependent on the electric field (whether it is uniform or nonuniform) and the location, whereas the electric potential energy also depends on the amount of charge the object in the system is experiencing. Also, although electric potential is an important topic to learn, most problems encountered will not ask to find just the "electric potential," instead, questions will most likely ask for the "electric potential difference." This is because electric potential is measured using different locations, or more specifically pathways between the different locations, so instead of determining the electric potential of location A and the electric potential of final location B, it would make more sense to determine the "difference in electric potential between locations A and B."

===A Mathematical Model===

Like mentioned in the Main Idea, instead of electric potential, in most cases, electric potential difference is needed to be found. The general equation for the potential difference is '''<math>∆{{U}_{electric}} = {q} * ∆{V} </math>'''.

'''<math>∆{{U}_{electric}}</math>''' is the electric potential energy, which is measured in Joules (J). '''q''' is the charge of the particle moving through the path of the electric potential difference, which is measured in coulombs (C). '''∆V''' is the electric potential difference, which is measured in Joules per Coulomb (J/C), or just Volts (V).

Aside from the general equation, the electric potential difference can also be found in other ways. The potential difference in an '''uniform field''' is '''<math>∆{V} = -({E}_{x}∆{x} + {E}_{y}∆{y} + {E}_{z}∆{z})</math>''', which can also be written as '''<math>∆{V} = -\vec{E}·∆\vec{l}</math>'''.

'''∆V''' is the electric potential difference, which is measured in Joules per Coulomb (J/C), or just Volts (V). '''E''' is the electric field, which is measured in Newtons per Coulomb (N/C), and it is important to note that the different direction components of the electric field are used in the equation. '''l''' (or the '''x''', '''y''', '''z''') is the distance between the two described locations, which is measured in meters, and x, y, and z, are the different components of the difference.

The electric potential difference in a '''nonuniform field''' is '''<math>∆{V} = -∑ \vec{E}·∆\vec{l}</math>'''. The different parts in this particular equation resembles the equation for the potential difference in an uniform field, except that with the nonuniform field, the potential difference in the different fields are summed up. This situation can be quite easy, but when the system gets difficult, first, choose a path and divide it into smaller pieces of '''<math>∆\vec{l}</math>'''; second, write an expression for '''<math>∆{V} = -\vec{E}·∆\vec{l}</math>''' of one piece; third, add up the contributions of all the pieces; last, check the result to make sure the magnitude, direction, and units make sense.

Aside from just calculating the value of the electric potential difference, determining the sign is also quite crucial to be successful. If the path being considered is in the same direction as the electric field, then the sign with be negative (-), or the potential is decreasing. If the path being considered is in the opposite direction as the electric field, then the sign will be positive (+), or the potential is increasing. If the path being considered is perpendicular to the electric field, then the potential difference will just be zero and have no direction. With these simple tips, the direction of the potential difference can be rechecked with the answer calculated using vectors.

Also, when working with different situations, it is nice to keep in mind that in a conductor, the electric field is zero. Therefore, the potential difference is zero as well. In an insulator, the electric field is '''<math>{E}_{applied} / K</math>''' where '''K''' is the dielectric constant. Also, the round trip potential difference is always zero, or in other words, if you start from a certain point and end at the same point, then, the potential difference will be zero.

===A Computational Model===
Click on the link to see Electric Potential through VPython!

Make sure to press "Run" to see the principle in action!

[https://trinket.io/embed/glowscript/0a7e486c94 Teach hand-on with GlowScript]

Watch this video for a more visual approach!

[https://www.youtube.com/watch?v=-Rb9guSEeVE Electric Potential: Visualizing Voltage with 3D animations]

==Examples==

===Simple===
In a capacitor, the negative charges are located on the left plate, and the positive charges are located on the right plate. Location A is at the left end of the capacitor, and Location B is at the right end of the capacitor, or in other words, Location A and B are only different in terms of their x component location. The path moves from A to B. What is the direction of the electric field? Is the potential difference positive or negative? [Hint: Draw a picture!]

'''Answer: The electric field is to the left. The potential difference is increasing, or is positive.'''

'''Explanation:'''
The electric field always moves away from the positive charge and towards the negative charge, which means the electric field in this example is to the left. Because the direction and the electric field and the direction of the path are opposite, the potential difference is increasing, or is positive.

===Middling===

Calculate the difference in electric potential between two locations A, which is at <-0.4, 0,0>m, and B, which is at <0.2,0,0>m. The electric field in the location is <500,0,0> N/C.

'''Answer: -300V'''

'''Explanation:'''

<math>∆\vec{l}</math> = <0.2,0,0>m - <-0.4,0,0>m = <0.6,0,0>m

<math>∆{V} = -({E}_{x}∆{x} + {E}_{y}∆{y} + {E}_{z}∆{z})</math>

<math>∆{V} = -(500 N/C * 0.6 m + 0*0 + 0*0)</math>

<math>∆{V} = -300 V</math>

===Difficult===

Suppose that from x=0 to x=3 the electric field is uniform and given by

'''Answer: -14.3 V ; 2.3e-18 J'''

'''Explanation:'''

<math>∆{V} = -\int_C^D {E}_{x} \, dx </math>

<math>∆{V} = -\int_{1e-10}^{2e-8} \tfrac{1}{4π{ε}_{0}}*\tfrac{Q}{{x}^{2}} \, dx </math>

<math>∆{V} = \tfrac{1}{4π{ε}_{0}}*{1.6e-19 C}*({\tfrac{1}{2e-8 m} - \tfrac{1}{1e-10 m}}) </math>

<math>∆{V}</math> = '''-14.3 V'''

∆K + ∆U = <math>{W}_{ext}</math>

<math>{W}_{ext}</math> = 0 + (-e)(∆V)

<math>{W}_{ext}</math> = (-1.6e-19 C)(-14.3 V)

<math>{W}_{ext}</math> = '''2.3e-18 J'''

==Connectedness==
'''How is this topic connected to something that you are interested in?'''

[A] I am interested in robotic systems and building circuit boards and electrical systems for manufacturing robots. While studying this section in the book, I was able to connect back many of the concepts and calculations back to robotics and the electrical component of automated systems.

[R] Since high school, I never really understood how to work with the voltmeter and what it measured, and I have always wanted to know, but although this particular wiki page did not go into the details and other branches of electric potential, it led me to find the answers to something I was interested in since high school, the concept of electric potential.

'''How is it connected to your major?'''

[A] I am a Mechanical Engineering major, so I will be dealing with the electrical components of machines when I work. Therefore, I have to know these certain concepts such as electric potential in order to fully understand how they work and interact.

[R] As a biochemistry major, electric potential and electric potential difference is not particularly related to my major, but in chemistry classes, we use electrostatic potential maps (electrostatic potential energy maps) that shows the charge distributions throughout a molecule. Although the main use in electric potential is different in physics and biochemistry (where physicists use it identify the effect of the electric field at a location), I still found it interesting as the concept of electric potential (buildup) was being used in quite a different way.

'''Is there an interesting industrial application?'''

[A] Electrical potential is used to find the voltage across a path. This is useful when working with circuit components and attempting to manipulate the power output or current throughout a component.

[R] Electric potential sensors are being used to detect a variety of electrical signals made by the human body, thus contributing to the field of electrophysiology.

==History==

The idea of electric potential, in a way, started with Ben Franklin and his experiments in the 1740s. He began to understand the flow of electricity, which eventually paved the path towards explaining electric potential and potential difference. Scientists finally began to understand how electric fields were actually affecting the charges and the surrounding environment. Benjamin Franklin first shocked himself in 1746, while conducting experiments on electricity with found objects from around his house. Six years later, or 261 years ago for us, the founding father flew a kite attached to a key and a silk ribbon in a thunderstorm and effectively trapped lightning in a jar. The experiment is now seen as a watershed moment in mankind's venture to channel a force of nature that was viewed quite abstractly.

By the time Franklin started experimenting with electricity, he'd already found fame and fortune as the author of Poor Richard's Almanack. Electricity wasn't a very well understood phenomenon at that point, so Franklin's research proved to be fairly foundational. The early experiments, experts believe, were inspired by other scientists' work and the shortcomings therein.

That early brush with the dangers of electricity left an impression on Franklin. He described the sensation as "a universal blow throughout my whole body from head to foot, which seemed within as well as without; after which the first thing I took notice of was a violent quick shaking of my body." However, it didn't scare him away. In the handful of years before his famous kite experiment, Franklin contributed everything from designing the first battery designs to establishing some common nomenclature in the study of electricity. Although Franklin is often coined the father of electricity, after he set the foundations of electricity, many other scientists contributed his or her research in the advancement of electricity and eventually led to the discovery of electric potential and potential difference.

== See also ==

Like mentioned multiple times throughout the page, although electric potential is a huge and important topic, it has many branches, which makes the concept of electric potential difficult to stand alone. Even with this page, to support the concept of electric potential, many crucial branches of the topic appeared, like potential difference (which also branched into [[http://www.physicsbook.gatech.edu/Potential_Difference_Path_Independence,_claimed_by_Aditya_Mohile Potential Difference Path Independence]], [[http://www.physicsbook.gatech.edu/Potential_Difference_in_a_Uniform_Field Potential Difference In A Uniform Field]], and [[http://www.physicsbook.gatech.edu/Potential_Difference_of_Point_Charge_in_a_Non-Uniform_Field Potential Difference In A Nonuniform Field]]).

===External Links===

[1] https://www.khanacademy.org/test-prep/mcat/physical-processes/electrostatics-1/v/electric-potential-at-a-point-in-space

[2] https://www.youtube.com/watch?v=pcWz4tP_zUw

[3] https://www.youtube.com/watch?v=Vpa_uApmNoo

==References==

[1] "Benjamin Franklin and Electricity." Benjamin Franklin and Electricity. N.p., n.d. Web. 17 Apr. 2016. <http://www.americaslibrary.gov/aa/franklinb/aa_franklinb_electric_1.html>.

[2] Bottyan, Thomas. "Electrostatic Potential Maps." Chemwiki. N.p., 02 Oct. 2013. Web. 17 Apr. 2016. <http://chemwiki.ucdavis.edu/Core/Theoretical_Chemistry/Chemical_Bonding/General_Principles_of_Chemical_Bonding/Electrostatic_Potential_maps>.

[3] "Electric Potential Difference." Electric Potential Difference. The Physics Classroom, n.d. Web. 14 Apr. 2016. <http://www.physicsclassroom.com/class/circuits/Lesson-1/Electric-Potential-Difference>.

[4] Harland, C. J., T. D. Clark, and R. J. Prance. "Applications of Electric Potential (Displacement Current) Sensors in Human Body Electrophysiology." International Society for Industrial Process Tomography, n.d. Web. 16 Apr. 2016. <http://www.isipt.org/world-congress/3/269.html>.

[5] Sherwood, Bruce A. "2.1 The Momentum Principle." Matter & Interactions. By Ruth W. Chabay. 4th ed. Vol. 1. N.p.: John Wiley & Sons, 2015. 45-50. Print. Modern Mechanics.

Electric Potential

2016-11-27T21:26:36Z

Zwright8:

'''AUTHOR: Rmohammed7'''

'''REVISED BY: HAYOUNG KIM (SPRING 2016)'''

"REVISED BY: ZACHARY WRIGHT ( FALL 2016)

This page discusses the Electric Potential and examples of how it is used.

----

== The Main Idea ==

Electric potential is a rather difficult concept as it is usually accompanied by other topics. For example, electric potential is not the same as electric potential energy as electric potential is defined as the total potential energy per charge and is basically used to describe the electric field's effect at a certain location. In other words, electric potential is purely dependent on the electric field (whether it is uniform or nonuniform) and the location, whereas the electric potential energy also depends on the amount of charge the object in the system is experiencing. Also, although electric potential is an important topic to learn, most problems encountered will not ask to find just the "electric potential," instead, questions will most likely ask for the "electric potential difference." This is because electric potential is measured using different locations, or more specifically pathways between the different locations, so instead of determining the electric potential of location A and the electric potential of final location B, it would make more sense to determine the "difference in electric potential between locations A and B."

===A Mathematical Model===

Like mentioned in the Main Idea, instead of electric potential, in most cases, electric potential difference is needed to be found. The general equation for the potential difference is '''<math>∆{{U}_{electric}} = {q} * ∆{V} </math>'''.

'''<math>∆{{U}_{electric}}</math>''' is the electric potential energy, which is measured in Joules (J). '''q''' is the charge of the particle moving through the path of the electric potential difference, which is measured in coulombs (C). '''∆V''' is the electric potential difference, which is measured in Joules per Coulomb (J/C), or just Volts (V).

Aside from the general equation, the electric potential difference can also be found in other ways. The potential difference in an '''uniform field''' is '''<math>∆{V} = -({E}_{x}∆{x} + {E}_{y}∆{y} + {E}_{z}∆{z})</math>''', which can also be written as '''<math>∆{V} = -\vec{E}·∆\vec{l}</math>'''.

'''∆V''' is the electric potential difference, which is measured in Joules per Coulomb (J/C), or just Volts (V). '''E''' is the electric field, which is measured in Newtons per Coulomb (N/C), and it is important to note that the different direction components of the electric field are used in the equation. '''l''' (or the '''x''', '''y''', '''z''') is the distance between the two described locations, which is measured in meters, and x, y, and z, are the different components of the difference.

The electric potential difference in a '''nonuniform field''' is '''<math>∆{V} = -∑ \vec{E}·∆\vec{l}</math>'''. The different parts in this particular equation resembles the equation for the potential difference in an uniform field, except that with the nonuniform field, the potential difference in the different fields are summed up. This situation can be quite easy, but when the system gets difficult, first, choose a path and divide it into smaller pieces of '''<math>∆\vec{l}</math>'''; second, write an expression for '''<math>∆{V} = -\vec{E}·∆\vec{l}</math>''' of one piece; third, add up the contributions of all the pieces; last, check the result to make sure the magnitude, direction, and units make sense.

Aside from just calculating the value of the electric potential difference, determining the sign is also quite crucial to be successful. If the path being considered is in the same direction as the electric field, then the sign with be negative (-), or the potential is decreasing. If the path being considered is in the opposite direction as the electric field, then the sign will be positive (+), or the potential is increasing. If the path being considered is perpendicular to the electric field, then the potential difference will just be zero and have no direction. With these simple tips, the direction of the potential difference can be rechecked with the answer calculated using vectors.

Also, when working with different situations, it is nice to keep in mind that in a conductor, the electric field is zero. Therefore, the potential difference is zero as well. In an insulator, the electric field is '''<math>{E}_{applied} / K</math>''' where '''K''' is the dielectric constant. Also, the round trip potential difference is always zero, or in other words, if you start from a certain point and end at the same point, then, the potential difference will be zero.

===A Computational Model===
Click on the link to see Electric Potential through VPython!

Make sure to press "Run" to see the principle in action!

[https://trinket.io/embed/glowscript/0a7e486c94 Teach hand-on with GlowScript]

Watch this video for a more visual approach!

[https://www.youtube.com/watch?v=-Rb9guSEeVE Electric Potential: Visualizing Voltage with 3D animations]

==Examples==

===Simple===
In a capacitor, the negative charges are located on the left plate, and the positive charges are located on the right plate. Location A is at the left end of the capacitor, and Location B is at the right end of the capacitor, or in other words, Location A and B are only different in terms of their x component location. The path moves from A to B. What is the direction of the electric field? Is the potential difference positive or negative? [Hint: Draw a picture!]

'''Answer: The electric field is to the left. The potential difference is increasing, or is positive.'''

'''Explanation:'''
The electric field always moves away from the positive charge and towards the negative charge, which means the electric field in this example is to the left. Because the direction and the electric field and the direction of the path are opposite, the potential difference is increasing, or is positive.

===Middling===

Calculate the difference in electric potential between two locations A, which is at <-0.4, 0,0>m, and B, which is at <0.2,0,0>m. The electric field in the location is <500,0,0> N/C.

'''Answer: -300V'''

'''Explanation:'''

<math>∆\vec{l}</math> = <0.2,0,0>m - <-0.4,0,0>m = <0.6,0,0>m

<math>∆{V} = -({E}_{x}∆{x} + {E}_{y}∆{y} + {E}_{z}∆{z})</math>

<math>∆{V} = -(500 N/C * 0.6 m + 0*0 + 0*0)</math>

<math>∆{V} = -300 V</math>

===Difficult===

Suppose that from x=0 to x=3 the electric field is uniform and given by

'''Answer: -14.3 V ; 2.3e-18 J'''

'''Explanation:'''

<math>∆{V} = -\int_C^D {E}_{x} \, dx </math>

<math>∆{V} = -\int_{1e-10}^{2e-8} \tfrac{1}{4π{ε}_{0}}*\tfrac{Q}{{x}^{2}} \, dx </math>

<math>∆{V} = \tfrac{1}{4π{ε}_{0}}*{1.6e-19 C}*({\tfrac{1}{2e-8 m} - \tfrac{1}{1e-10 m}}) </math>

<math>∆{V}</math> = '''-14.3 V'''

∆K + ∆U = <math>{W}_{ext}</math>

<math>{W}_{ext}</math> = 0 + (-e)(∆V)

<math>{W}_{ext}</math> = (-1.6e-19 C)(-14.3 V)

<math>{W}_{ext}</math> = '''2.3e-18 J'''

==Connectedness==
'''How is this topic connected to something that you are interested in?'''

[A] I am interested in robotic systems and building circuit boards and electrical systems for manufacturing robots. While studying this section in the book, I was able to connect back many of the concepts and calculations back to robotics and the electrical component of automated systems.

[R] Since high school, I never really understood how to work with the voltmeter and what it measured, and I have always wanted to know, but although this particular wiki page did not go into the details and other branches of electric potential, it led me to find the answers to something I was interested in since high school, the concept of electric potential.

'''How is it connected to your major?'''

[A] I am a Mechanical Engineering major, so I will be dealing with the electrical components of machines when I work. Therefore, I have to know these certain concepts such as electric potential in order to fully understand how they work and interact.

[R] As a biochemistry major, electric potential and electric potential difference is not particularly related to my major, but in chemistry classes, we use electrostatic potential maps (electrostatic potential energy maps) that shows the charge distributions throughout a molecule. Although the main use in electric potential is different in physics and biochemistry (where physicists use it identify the effect of the electric field at a location), I still found it interesting as the concept of electric potential (buildup) was being used in quite a different way.

'''Is there an interesting industrial application?'''

[A] Electrical potential is used to find the voltage across a path. This is useful when working with circuit components and attempting to manipulate the power output or current throughout a component.

[R] Electric potential sensors are being used to detect a variety of electrical signals made by the human body, thus contributing to the field of electrophysiology.

==History==

The idea of electric potential, in a way, started with Ben Franklin and his experiments in the 1740s. He began to understand the flow of electricity, which eventually paved the path towards explaining electric potential and potential difference. Scientists finally began to understand how electric fields were actually affecting the charges and the surrounding environment. Benjamin Franklin first shocked himself in 1746, while conducting experiments on electricity with found objects from around his house. Six years later, or 261 years ago for us, the founding father flew a kite attached to a key and a silk ribbon in a thunderstorm and effectively trapped lightning in a jar. The experiment is now seen as a watershed moment in mankind's venture to channel a force of nature that was viewed quite abstractly.

By the time Franklin started experimenting with electricity, he'd already found fame and fortune as the author of Poor Richard's Almanack. Electricity wasn't a very well understood phenomenon at that point, so Franklin's research proved to be fairly foundational. The early experiments, experts believe, were inspired by other scientists' work and the shortcomings therein.

That early brush with the dangers of electricity left an impression on Franklin. He described the sensation as "a universal blow throughout my whole body from head to foot, which seemed within as well as without; after which the first thing I took notice of was a violent quick shaking of my body." However, it didn't scare him away. In the handful of years before his famous kite experiment, Franklin contributed everything from designing the first battery designs to establishing some common nomenclature in the study of electricity. Although Franklin is often coined the father of electricity, after he set the foundations of electricity, many other scientists contributed his or her research in the advancement of electricity and eventually led to the discovery of electric potential and potential difference.

== See also ==

Like mentioned multiple times throughout the page, although electric potential is a huge and important topic, it has many branches, which makes the concept of electric potential difficult to stand alone. Even with this page, to support the concept of electric potential, many crucial branches of the topic appeared, like potential difference (which also branched into [[http://www.physicsbook.gatech.edu/Potential_Difference_Path_Independence,_claimed_by_Aditya_Mohile Potential Difference Path Independence]], [[http://www.physicsbook.gatech.edu/Potential_Difference_in_a_Uniform_Field Potential Difference In A Uniform Field]], and [[http://www.physicsbook.gatech.edu/Potential_Difference_of_Point_Charge_in_a_Non-Uniform_Field Potential Difference In A Nonuniform Field]]).

===External Links===

[1] https://www.khanacademy.org/test-prep/mcat/physical-processes/electrostatics-1/v/electric-potential-at-a-point-in-space

[2] https://www.youtube.com/watch?v=pcWz4tP_zUw

[3] https://www.youtube.com/watch?v=Vpa_uApmNoo

==References==

[1] "Benjamin Franklin and Electricity." Benjamin Franklin and Electricity. N.p., n.d. Web. 17 Apr. 2016. <http://www.americaslibrary.gov/aa/franklinb/aa_franklinb_electric_1.html>.

[2] Bottyan, Thomas. "Electrostatic Potential Maps." Chemwiki. N.p., 02 Oct. 2013. Web. 17 Apr. 2016. <http://chemwiki.ucdavis.edu/Core/Theoretical_Chemistry/Chemical_Bonding/General_Principles_of_Chemical_Bonding/Electrostatic_Potential_maps>.

[3] "Electric Potential Difference." Electric Potential Difference. The Physics Classroom, n.d. Web. 14 Apr. 2016. <http://www.physicsclassroom.com/class/circuits/Lesson-1/Electric-Potential-Difference>.

[4] Harland, C. J., T. D. Clark, and R. J. Prance. "Applications of Electric Potential (Displacement Current) Sensors in Human Body Electrophysiology." International Society for Industrial Process Tomography, n.d. Web. 16 Apr. 2016. <http://www.isipt.org/world-congress/3/269.html>.

[5] Sherwood, Bruce A. "2.1 The Momentum Principle." Matter & Interactions. By Ruth W. Chabay. 4th ed. Vol. 1. N.p.: John Wiley & Sons, 2015. 45-50. Print. Modern Mechanics.