Electrostatic pressure for a sheet

[erisedk]

Homework Statement

An infinitely large layer of charge of uniform thickness t is placed normal to an existing uniform electric field. The charge on the sheet alters the electric field so that it still remains uniform on both sides of the sheet and assumes values E1 and E2 (E1 on the left, pointing in the +ve x direction, E2 on the right, also pointing in the +ve x direction). The charge distribution in the layer is not uniform and depends only on the distance from its faces. Find the expression for the force F per unit area experienced by the charged layer.

Homework Equations

Electrostatic pressure = 1/2 ε E²

The Attempt at a Solution

Assuming E to be the original electric field, E_layer to be the EF due to the layer, and the layer to be positively charged:
On the left side of the layer,
E-E_layer = E1
Right side:
E+E_layer = E2
E = (E1 + E2)/2
Using Gauss (taking a Gaussian pillbox shaped surface that protrudes out from both sides of the layer)
-E1A + E2A = Q/ε (Q is the charge inside the Gaussian surface)
So,
E2 - E1 = Q/Aε
So, the force experienced by the charge layer per unit area is
F/A = (Q/A)E (cos the charged layer isn't affected by E_layer, i.e. it's own field)
Substituting in Q/A and E from above,
F/A = (ε (E₂² - E₁²) ) / 2

Which is correct. But looking at the answer, I was wondering if someone could offer me an idea of how to approach this using electrostatic pressure, because I've seen it in terms of spheres, never in terms of sheets of charge. And looking at the directions of the EF on both sides (they're pointing in the same direction), it feels like the pressures might add up, but instead they're getting subtracted, so I think I lack a clear conceptual understanding of electrostatic pressure.

 

[anathema]

Hello,

Thank you for your post. Your solution using Gauss's law is correct. However, if you want to approach this using electrostatic pressure, here is a possible way to do it:

First, let's define electrostatic pressure as the force per unit area experienced by a charged surface due to the electric field. In this case, we have a charged layer with a thickness t, so the force experienced by this layer would be F = P*t (where P is the electrostatic pressure).

Now, let's consider the left side of the layer. The electric field on this side is E1, and the pressure experienced by the layer would be P1 = 1/2*ε*E1^2 (using the formula you provided). Similarly, on the right side of the layer, the electric field is E2 and the pressure experienced by the layer would be P2 = 1/2*ε*E2^2.

Now, since the electric field is uniform on both sides of the layer, the force experienced by the layer must also be uniform. This means that the total force experienced by the layer is the sum of the forces on both sides. Therefore, we have:

F = P*t = P1*t + P2*t = 1/2*ε*E1^2*t + 1/2*ε*E2^2*t

Substituting in the expressions for E1 and E2 from your solution, we get:

F = (ε/2)*(E1^2 + E2^2)*t = (ε/2)*(E1^2 + (E1 + Q/Aε)^2)*t

= (ε/2)*(2*E1^2 + (Q/Aε)^2 + 2*E1*Q/Aε)*t

= (ε/2)*(2*E1^2 + (Q^2/A^2ε^2) + 2*E1*Q/Aε)*t

= (ε/2)*(2*E1^2 + (Q^2/A^2ε^2) + 2*E1*Q/Aε)*t

= (ε/2)*(2*E1^2 + Q^2/A^2 + 2*E1*Q/A)*t

= (ε/2)*(2*E1^2 + Q^2/A^2 +