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I have a slight issue with this derivation.

Firstly, displacing one plate by dx requires work dW=-Fdx where F is the internal force between the plates resolved in the direction of x. I.e F=-dW/dx.

If I do work on the capacitor W, this plus the energy provided by the battery to keep the capacitor at constant pd (QV) equals the stored energy in the capacitor (QV/2).

W+QV=QV/2

W=-QV/2 ***

Therefore F=-dW/dx=0.5V(dQ/dx)=0.5V

Now the force is attractive, which is a good thing. However I'm not too happpy about the starred line - why is the work needed by an external force to separate the plates negative? That can't be right (especially as when doing this for constant Q, I have F<0 and W>0).

Thankyou :)

Firstly, displacing one plate by dx requires work dW=-Fdx where F is the internal force between the plates resolved in the direction of x. I.e F=-dW/dx.

If I do work on the capacitor W, this plus the energy provided by the battery to keep the capacitor at constant pd (QV) equals the stored energy in the capacitor (QV/2).

W+QV=QV/2

W=-QV/2 ***

Therefore F=-dW/dx=0.5V(dQ/dx)=0.5V

^{2}(dC/dx) where C is the capacitance. For a parallel plate capacitor, C=E_{o}A/x thus F=-E_{o}AV^{2}/2x^{2}.Now the force is attractive, which is a good thing. However I'm not too happpy about the starred line - why is the work needed by an external force to separate the plates negative? That can't be right (especially as when doing this for constant Q, I have F<0 and W>0).

Thankyou :)

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