What Is the Potential Difference Between Two Charged Parallel Plates?

In summary, the potential difference between two parallel plates with equal but opposite charges of magnitude 36 nC/m^2 and separated by .12 m can be found by using the formula Delta V = -ESds, where E is the electric field, sigma is the surface charge density, and A is the area of the plate. By using Gauss' law and creating a cylindrical gaussian surface, the electric field can be found and substituted into the formula to determine the potential difference.
  • #1
jordanl122
14
0
Two parallel plates having charges of equal magnitude but opposite sign are separated by .12 m. Each plate has a surface charge density of 36 nC/m^2. Determine the potential difference between the plates.

I have to admit I am at a bit of a loss on how to answer this
(lettings S represent the integral sign)
I know Delta V = -SEds
and that E is constant so
Delta V = -ESds so delat V = -Ed
but I am not sure how to find E, I know sigma =Q/A
and I don't see how Gauss' law could really work here

any light someone can shed would be greatly appreciated.
 
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  • #2
Pretty classic problem. Assume the plate is infinite and make a cylindrical gaussian surface whose length is perpendicular to the face of a sheet: Illustrated in better detail here

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elesht.html#c2 Just remember that through the top and bottom of the cylinder flux equal to EA comes out, so the total flux is 2EA.
 
  • #3


The potential difference between the two plates can be determined by using the formula Delta V = Ed, where E is the electric field between the plates and d is the distance between them. In this case, we can use Gauss's law to calculate the electric field between the plates.

Gauss's law states that the electric flux through a closed surface is equal to the charge enclosed by that surface divided by the permittivity of free space (ε0). In this problem, we can imagine a closed cylinder with its axis perpendicular to the plates and passing through the center of the plates. The electric flux through this cylinder is equal to the charge enclosed by it, which is the sum of the charges on each plate. Thus, we can write the equation:

Φ = Qenc/ε0 = (σA + σA)/ε0 = 2σA/ε0

where Φ is the electric flux, Qenc is the enclosed charge, σ is the surface charge density, and A is the area of one plate.

Since the electric field is perpendicular to the plates, the electric flux is also equal to the electric field multiplied by the area of the plates (A). Therefore, we can write:

Φ = EA

Equating the two equations for electric flux, we get:

EA = 2σA/ε0

Solving for E, we get:

E = 2σ/ε0

Substituting the given values, we get:

E = 2(36 nC/m^2)/(8.85 x 10^-12 C^2/Nm^2) = 8.1 x 10^9 N/C

Now, we can use the formula Delta V = Ed to calculate the potential difference between the plates:

Delta V = (8.1 x 10^9 N/C)(0.12 m) = 9.72 x 10^8 V

Therefore, the potential difference between the plates is 9.72 x 10^8 V.
 

FAQ: What Is the Potential Difference Between Two Charged Parallel Plates?

What is potential difference?

Potential difference refers to the difference in electrical potential energy between two points in an electric circuit. It is also known as voltage and is measured in volts (V).

How is potential difference calculated?

Potential difference is calculated by dividing the change in electrical potential energy by the charge, or by multiplying the electric field strength by the distance between two points. The formula for potential difference is V = W/Q, where V is potential difference, W is work, and Q is charge.

What are the units of potential difference?

The units of potential difference are volts (V), named after Italian physicist Alessandro Volta. In some contexts, potential difference may also be measured in joules per coulomb (J/C).

What causes potential difference?

Potential difference is caused by the presence of an electric field, which is created by the separation of positive and negative charges. The magnitude of the potential difference is determined by the strength of the electric field.

What are some real-life applications of potential difference?

Potential difference has many practical applications, including powering electronic devices, such as cell phones and laptops, and providing electricity for lighting and heating. It is also used in medical equipment, such as MRI machines, and in transportation systems, such as electric cars.

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