A parallel-plate vacuum capacitor with plate area A and separation x has charges

Question

A parallel-plate vacuum capacitor with plate area A and separation x has charges +Q and -Q on its plates. The capacitor is disconnectedfrom the source of the charge, so the charge on each plate remains fixed. (a) What is the total energy stored in the capacitor? (b) The plates are pulled apart an additional distance dx. What is the change in the stored energy? (c) if F is the force with which the plates attract each other,then the change in the stored energy must be equal to work dW=Fdx done in pulling the plates apart. Find an expression for F

Explanation / Answer

C=A/x

ENERGY =Q^2/2C=Q^2/2*A/x=(Q^2*x)/2*A

change in enrgy when pulled by dx=Q^2dx/2A

Fdx=Q^2dx/2A

cancelling dx

F=Q^2/2A

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