What Is the Potential Difference Between Two Charged Parallel Plates?

[jordanl122]

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.

 

[vsage]

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.

 

[wais]

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.