# Homework Help: Work to separate a plane capacitor

1. Jun 15, 2017

### Kelly Lin

1. The problem statement, all variables and given/known data
A capacitor with C is charged by a battery to a voltage V and then disconnected. The distance between plates is slowly increased by an external force. What is the change of the electrostatic energy of capacitor during this process?

2. Relevant equations
I have two ways to solve the problem but I don't know which is correct.

3. The attempt at a solution
If the distance changed from d1 to d2, then
$$C'=C\frac{d_{1}}{d_{2}} \\ V'=V\frac{d_{2}}{d_{1}} \\ \frac{1}{2}C'V'^{2}-\frac{1}{2}CV^{2}=\frac{1}{2}(C\frac{d_{1}}{d_{2}})(V\frac{d_{2}}{d_{1}})^{2}=\frac{1}{2}CV^{2}(\frac{d_{1}}{d_{2}}-1)$$
On the other hand, work can be defined as W=Q[V'-V], then
$$W=Q(V'-V)=QV(\frac{d_{1}}{d_{2}}-1)=CV^{2}(\frac{d_{1}}{d_{2}}-1)$$
But it seems that there is a missing factor (1/2) in the second solution!
Where did I go wrong?

2. Jun 15, 2017

### BvU

Here already :$$\frac{1}{2}C'V'^{2}-\frac{1}{2}CV^{2}=\frac{1}{2}(C\frac{d_{1}}{d_{2}})(V\frac{d_{2}}{d_{1}})^{2}=\frac{1}{2}CV^{2}(\frac{d_{1}}{d_{2}}-1)$$W changes from + to - halfway ?
And try $dW = Q dV$ for the second approach

3. Jun 15, 2017

### Kelly Lin

Okay!! I got it!!
Thanks a lot! haha!