How Is Work Calculated When Separating Charged Sheets?

[NullSpace0]

Homework Statement

I have a current problem set question about the work per unit area required to separate infinite sheets of charge with equal and opposite charge densities from a separation of d to a separation of 2d.

Homework Equations

U=(1/8∏)∫E²dV
W=∫F*dr
E=4∏σ

The Attempt at a Solution

I was thinking I could just find the difference in energy stored in the field before and after... so I would integrate E² over the initial and final volumes, and then the difference must come from work I have put into the system, which shows up as energy stored in the electric field.

If I do this, I get U_i=2∏σ²d and U_f=4∏σ²d

And so the work is just 2∏σ²d. But how am I sure that this has units work per area? It seems like work per volume because you have (esu^2)/(cm^3) for the units written out fully.

Is there a way to do this by integrating the force (or perhaps the field) rather than finding stored energy changes?

 

[vela]

Remember the potential energy of two charges is $U = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r}$ (in SI), so energy is charge²/length.

 

[schaefera]

I believe the answer is correct.

 

[NullSpace0]

What about the force method? Or any other method that's valid?

 

[schaefera]

I actually don't know how to do any other method- ideas from other people?

 

[TSny]

I think you can argue that the force felt by a patch of area dA on one sheet is due solely to the electric field produced by the other sheet (E = 2∏σ). The force on the patch is then dq*E where dq = σdA. Since the force will be constant as you separate the plates, the work will just be F*d. I believe this will give you the same answer as the energy method.