# Capacitance and E-field of a parallel plate capacitor

1. Feb 22, 2013

### unscientific

1. The problem statement, all variables and given/known data

3. The attempt at a solution

I know that the E-field is definitely affected by the presence of the space charges. Since one side is positive, the positive charges in the air get pushed towards the negative plate.

so

Enet = Eplate - Echarges

But the thing is I don't know how the distribution of the charges will end up as, and hence how to calculate Enet.

2. Feb 22, 2013

### ehild

The distribution of the space charge does not change, the charges do not move.

ehild

3. Feb 22, 2013

### unscientific

Then surely the E-field would be σ/ε0?

But in the answer, clearly the E-field depends on x.

4. Feb 22, 2013

### jaumzaum

The charges do not move. We clearly have an E field between the 2 plates due to the charge in the middle. Considering V=0 , we can see the electric field inside would be zero in the middle, positive in one side and negative in the other. This way -$\int \vec{E} \vec{dx}$ = ΔV = 0, so the charges do not affect the voltage of the capacitor, as well as its capacitance. Now considering V≠0, by the superposing theorem we can add the V/d to the electric field due to the charges in the middle (this electric field will only be zero in x=d/2).

5. Feb 22, 2013

### unscientific

I see what you mean..The problem here is that I'm trying to think about how the space charges are distributed and what the E-field due to them would be.

Let's just take all space charge to be positive Q. Thus ρ = Q/V
At midway there are equal amount of positive charge on top and bottom, so the E-field strength would be equal and cancel.

But as you move closer to the bottom, there are more positive charges above.

But I still do not know how the E-field depends on the number of charges and distance...We could model them as a solid rectangular block whose E-field is only perpendicular to the surface but the question remains: How does the E-field strength change with distance?

This question is under the topic of Capacitance...

6. Feb 22, 2013

### jaumzaum

The charges are uniformly distributed. Consider each rectangle (as you sad) to be a positive plate of a capacitor. We know the E field due to a infinite plate is σ/2ε0. But the σ of each of them would be different right?
How can we write σ in function of ρ, x and d?

7. Feb 22, 2013

### unscientific

Ok i think i got the answer!

Take a thin rectangular block of thickness dx and area A.

Amount of charge in that block=
dQ = p (dV)

σ = (dQ)/A = (p/A)dV = p dx

dE = σ/2ε = (p/2ε) dx

Thus E = integral from 0 to d-x of dE - integral from 0 to x = (p/2ε)(d-x) + (p/2ε)(-x) = (p/2ε)(d-2x)

as bottom the E-field points upwards while at the top the E-field points downwards

Last edited: Feb 22, 2013