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)=0.5V2(dC/dx) where C is the capacitance. For a parallel plate capacitor, C=EoA/x thus F=-EoAV2/2x2.
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)=0.5V2(dC/dx) where C is the capacitance. For a parallel plate capacitor, C=EoA/x thus F=-EoAV2/2x2.
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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