Potential Difference of Point Charge in a Non-Uniform Field

Claimed by dmengesha3 Dina

The Main Idea

Non uniform electric fields are generated by point charges because the magnitude is different depending on the location at which the electric field is being observed (refer to the equation of the electric field made by a point charge). If one is trying to find the electric potential between two different locations due to a point charge the difference in electric fields must be taken into account. Generally, in a non-uniform electric field one can write the change in potential difference between two locations to be: [math]\displaystyle{ \textstyle\int\limits_{i}^{f}-Edl }[/math] -- i being the intial location and f being the final location.

A Mathematical Model

In order to apply this general formula for potential difference to a point charge:

We substitute E for the formula for the electric field of a point charge and dl for r the distance between the point charge and the observation location.

[math]\displaystyle{ \textstyle\int\limits_{i}^{f}(9x10^9)\frac{Q}{r^2}dr }[/math]

If we carry out the integration we find that Vf- Vi = [math]\displaystyle{ \vartriangle(9x10^9)\frac{Q}{r} }[/math] which can also be written [math]\displaystyle{ ((9x10^9)\frac{Q}{rf})-((9x10^9)\frac{Q}{ri}) }[/math]

This equation can be used to determine the potential difference of a point charge between any two different locations.

A Computational Model

Refer to this Website:

https://phet.colorado.edu/sims/charges-and-fields/charges-and-fields_en.html

To observe how the electric field of a point charge differs from one location to another because it is proportional to 1/r^2. Then play around with measuring the potential difference between two points and see how distance affects the calculation of potential difference. You should observe a 1/r relationship.

Exercises 1. Put down a positive charge. Put 3 E-Field Sensors down: one close to the charge, the next farther away, the next one even farther stil

Notice that the magnitude of the arrows decreases as the distance from the point charge increases

2. Repeat the steps of #1 but replace the positive point charge with a negative point charge.

You should notice the same trend occur

3. Click "Show E-field" on the right menu bar and move the point charge around the plot

Notice that the Electric field gets larger in magnitude as the point charge moves closer to the observation location

4. Move the point charge closer to the observation location that measures potential difference

Notice that the closer the point charge is to this location the higher the magnitude of the potential difference

Examples

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

Simple

Refer to the figure above to answer the following question:

If the path is along the line i to f what direction is the electric field of the dipole?

down up left right Cannot determine

Answer: To the left. The electric field would point toward the negative end of the dipole

Is the potential difference Vf-Vi positive negative zero

Answer: It is negative because both deltal (displacement) and Electric field are pointing in the same direction which means that their product would have been positive and the negative sign in front of the expression -Edl would make the answer negative.

Shortcut to remember: E and dl same direction --> negative potential difference E and dl opposite direction-->positive potential difference

Middling

Calculate the change in potential moving from location [math]\displaystyle{ (9x10^-9) m }[/math] to location [math]\displaystyle{ (9.7x10^-9) m }[/math] from a proton.

Working it through

the final location is: (9.7x10^-9) m

the intial location is: (9x10^-9) m

So if we plug these values into the above equation we found by integrating Edr.

We have: [math]\displaystyle{ ((9x10^9)\frac{1.6x10^-19}{9.7x10^-9})-((9x10^9)\frac{1.6x10^-19}{9x10^-9}) }[/math]

Which yields a potential difference of -0.0115 V.

Difficult

The potential calculated for a point charge in a particular region of space is written by V=14xy+6y-4z, what are the electric field components at location <x,y,z>?

Working it through

Remember that potential difference [math]\displaystyle{ \textstyle\int\limits_{i}^{f}-Edl }[/math] thus Electric field can be calculated as the [math]\displaystyle{ \frac {-dv}{dx} }[/math] So one would take the derivative of the potential difference with respect to each variable giving:

E= <-14y,-14x-6,4>

Connectedness

Connections in Chemistry

Electrostatic potential energy maps are made of molecules to portray the charge distribution of a molecule 3 dimensionally. These maps can be used to determine electronegativity, bond characteristics, and also help find the reactive sites of molecules. Reactive sites are defined as a charged region of a molecule that interact with other charged particles. This can become especially important in assessing what types of molecular interactions and reactions will take place between two elements or compounds.

Industrial Application

Coulomb Barrier for Nuclear Fusion: For two particles (ex: 2 protons to fuse), they must be able to get close enough to one another for the nuclear strong force to overcome their electric repulsion. One must understand the potential difference of a point charge and the potential energy that creates a barrier between two point charges which can be calculated using U=ke^2/r where k =9e9 and e=1.6e-19 The r calculated using this formula determines the radius at which the nuclear attractive force becomes dominant.

History

In 1800, Alessandro Volta (pictured above) found that metals such as zinc and copper could produce currents; thus, he constructed the first battery what is know called a voltic pile. It was constructed from pieces of zinc and copper in salt water which produced an electric current. The SI unit for potential difference is named after Alessandro Volta.

Refer to:

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

To read more about Volta's early works and life.