Potential Difference in a Uniform Field: Difference between revisions

Revision as of 01:42, 27 November 2016

By: dachtani3

CLAIMED BY SHAN SUEN (FALL 2016)

Electric potential is a scalar quantity that is used to describe the change in electric potential energy per unit charge. This page will elaborate on the change in electric potential in a uniform field.

The Main Idea

Potential difference is the change in electric potential between a final and initial location when work is done on a charge to to affect its potential energy. The unit for electric potential is a volt (which is also joule/coulomb).

A Mathematical Model

This equation represents potential difference, where V is electric potential, U is electric potential energy, and q is unit charge.

[math]\displaystyle{ \Delta V }[/math] = [math]\displaystyle{ \frac{dU}{q} }[/math]

You can rearrange this equation to also show that electric potential energy is electric potential times unit charge.

[math]\displaystyle{ \Delta U }[/math] = [math]\displaystyle{ q \Delta V }[/math]

In addition, you can express electric potential as the sum of the dot product of electric field and displacement in each dimension. This expression is as follows:

[math]\displaystyle{ \Delta V }[/math] = -[math]\displaystyle{ \vec{E} }[/math]●[math]\displaystyle{ \Delta \vec{l} }[/math] = -(Ex●[math]\displaystyle{ \Delta x }[/math] + Ey●[math]\displaystyle{ \Delta y }[/math] + Ez●[math]\displaystyle{ \Delta z }[/math])

You can also use dot product notation:

[math]\displaystyle{ \Delta V }[/math] = -[math]\displaystyle{ \vec{E} }[/math]●[math]\displaystyle{ \Delta \vec{l} }[/math] = [math]\displaystyle{ \-\lt E_{x}, E_{y}, E_{z}\gt }[/math]

Potential difference can be either positive or negative because each component of electric field and displacement can be positive or negative. Within this expression, you can also note that the units for the electric field are V/m. We originally learned that units for electric field are N/c, but V/m is also an appropriate unit.

A Computational Model

-A good video that showcases 3D-models of the interaction of charges with electric potential and electric potential energy can be found here: https://www.youtube.com/watch?v=-Rb9guSEeVE

-In addition, this image showcases the relationship between electric fields and electric potential. If you take the negative gradient of electric potential, the result is the electric field. The gradient is the direction of the hills seen in the graph representing electric potential.

Examples

Simple

Questions 1 and 2 are based on the following situation:

A path consists of two locations. Location 1 is at <.1,0,0>m and Location 2 is at <.5,0,0>m. A uniform electric field of <300,0,0> N/C exists in this region pointing from Location 1 to Location 2.

Question 1: What is the difference in electric potential between the two points?

dl = final location - initial location = <.5,0,0> - <.1,0,0> = <.4,0,0> m

dV = -(Exdx + Eydy + Ezdz) = -(300 N/C*.4 + 0*0 + 0*0) = -120 V

Answer: -120 V

Question 2: What is the change in electric potential energy for a proton on this path?

dU = dV*q = -120 V * 1.6e-19 C = -1.92e-17 J

Answer: - 1.92e-17 J

Middling

Key Point to remember about signs:

-If the path is going in the direction of the electric field, electric potential is decreasing.

Ex: Refer to Simple Example

-If the path is going in the opposite direction of the electric field, electric potential is increasing.

Question: Location 1 is at <.1,0,0> and Location 2 is at <.5,0,0>. The uniform electric field is <300,0,0> N/C. The charge travels from Location 2 to Location 1.

new dl = <.1,0,0> - <.5,0,0> = <-.4,0,0>m

dV = -Exdx = -(-.4*300) = +120V

Answer: + 120 V

-If the path is perpendicular to the electric field, electric potential does not change.

Question: Location 1 is at <.1,0,0> and Location 2 is at <.5,0,0>. The uniform electric field is <0,300,0> N/C. The charge travels from Location 1 to Location 2.

new dL = <.5,0,0> - <.1,0,0> = <.4,0,0>m

dV = -(Exdx + Eydy + Exdz) = -(.4*0 + 0*300 +0*0) = 0 V

Answer: 0V

Difficult

The difficult example will refer to using a path that is not directly parallel to the electric field.

Question: Assume an electric field that has a magnitude of 300 N/C. The electric field is uniform. The path chosen is 5 m, 50 degrees away from the field. What is the difference in electric potential in this situation?

dV = -E*dl = -E*l*cos(theta)

dV = -(300 N/C)*(5 m)*cos(50) = -964 V

Answer: -964 V

Connectedness

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

Electric potential is related to something that I am interested in because of an internship I had dealing with circuit boards. Recently, I interned in the electronics systems lab at the Georgia Tech Research Institute. Our project was to build a integrated circuit board that acting as a no-parking zone sign for illegal parking areas. My particular board sensed the car, held the signal, and then transferred the electric signal to the audio board to display noise. On my board, I had many electronic components, including capacitors. During my internship, I placed the appropriate pieces on the board where deemed necessary, but I never really understood the physics of electricity behind the components. Hence, learning about electric potential now gives me a greater scope as to understand how electric charge and potential difference affects capacitors.

How is it connected to your major?

I am an Industrial & Systems engineering major. My major is all about understanding high level processes in order to identify areas for improvement. In particular, we have the opportunity to take intro to Electrical & Computer Engineering courses to understand the ins and outs of computers and electricity processes. We learn the basics of electricity, including the history of electric scientists like Volta, as well as the definitions of electric potential and electric potential energy. These intro level classes give us a strong foundation for upper-level classes where processes we encounter become very mathematically complex.

Is there an interesting industrial application?

An interested industrial application of electric potential is the ability to use electric potential sensors in human body electrophysiology. In a study conducted by the Center for Physical Electronics and Quantum Technology, the team utilized these sensors to detect electric signals in the human body. The sensors were mainly built with electrometer amplifiers. Specific to electric potential, the senors focus on displacement current rather than electric current at certain locations. The following signals were detected: electrocardiograms, electroencephalograms, and electro-oculograms, and the sensors were able to find the three signals without directly touching the human body. With these findings, the scientists are able to create "spatio-temporal array imaging" of different areas of the human body, including the heart and brain!

History

-Alessandro Volta, and Italian physicist, contributed many ideas and inventions to the field of electricity. He invented the first electric battery, the first electromotive series, and most notably, contributed to the idea of electric potential and its unit, the volt.

-In 1745, Volta was born. He spent most of his childhood experimenting with electricity in his friend's physics lab. When he was 18, he started communicating with physicists, Jean-Antonie Nollet and Giambatista Beccaria, who encouraged him to continue with his experiments.

-In 1775, Volta began teaching physics. He soon was able to isolate methane gas, which he discovered could produce electric sparks. In 1776, Volta put the two ideas together to conclude that he could send electric signals across Italy with the sparking machine.

-In 1778, Volta discovered electric potential, or voltage. He realized that the electric potential in a capacitor is directly proportional to the electric charge in that capacitor.

-In 1800, Volta combined all of his findings to create the voltaic pile, or the first electrochemical cell. This battery made of zinc and copper was able to produce a steady and constant electric current.

-Batteries today serve as the major practical application of electric potential. The unit for electric potential, the Volt, is named after Alessandro Volta and his contributions to the field of electricity.

@@ Line 19: / Line 19: @@
 In addition, you can express electric potential as the sum of the dot product of electric field and displacement in each dimension. This expression is as follows:
 :<math>\Delta V</math> = -<math>\vec{E}</math>●<math>\Delta \vec{l}</math> = -(Ex●<math>\Delta x</math> + Ey●<math>\Delta y</math> + Ez●<math>\Delta z</math>)
+You can also use dot product notation:
+:<math>\Delta V</math> = -<math>\vec{E}</math>●<math>\Delta \vec{l}</math> = <math>\-<E_{x}, E_{y}, E_{z}> </math>

Potential Difference in a Uniform Field: Difference between revisions

Revision as of 01:42, 27 November 2016

Contents

The Main Idea

A Mathematical Model

A Computational Model

Examples

Simple

Middling

Difficult

Connectedness

History

See also

Further reading

External links

References

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