Work to separate a plane capacitor

[Kelly Lin]

Homework Statement

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?

Homework Equations

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

The Attempt at a Solution

If the distance changed from d₁ to d₂, then
<br /> C&#039;=C\frac{d_{1}}{d_{2}}<br /> \\<br /> V&#039;=V\frac{d_{2}}{d_{1}}<br /> \\<br /> \frac{1}{2}C&#039;V&#039;^{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)<br />
On the other hand, work can be defined as W=Q[V'-V], then
<br /> W=Q(V&#039;-V)=QV(\frac{d_{1}}{d_{2}}-1)=CV^{2}(\frac{d_{1}}{d_{2}}-1)<br />
But it seems that there is a missing factor (1/2) in the second solution!
Where did I go wrong?

 

[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

 

[Kelly Lin]

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

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