a parallel-plate capacitor with plate are A and separation x has charges +Q and

Question

a parallel-plate capacitor with plate are A and separation x has charges +Q and -Q on its plates. The capacitor is disconnected from the source of 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 equal to work dW= Fdx done in pulling the plates apart. Find an expression for F

Explanation / Answer

capacitance = *A/x

a) E=Q^2/(2*C) = (Q^2)*x/(2**A)
b) new capacitance = *A/(x+dx)

thus E' = (Q^2)*(x+dx)/(2**A)

c) E = E' - E = (Q^2)*(dx)/(2**A) = F*dx

=> F = (Q^2)/(2**A)

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