A parallel-plate capacitor carries a constant charge Q (i.e., +Q on one plate, a

Question

A parallel-plate capacitor carries a constant charge Q (i.e., +Q on one plate, and -Q on the other). When a dielectric is inserted between the plates, what happens to the voltage difference between the plates, and the potential energy stored in the capacitor? A parallel-plate capacitor is attached to a battery that maintains a constant voltage difference ?V across the plates. When a dielectric is inserted between the plates, what happens to the charge on the plates, and the potential energy stored in the capacitor?

Explanation / Answer

capacitance=C=A/d

if dielectric is inserted then

C'=kA/d

k=dielectric constant

Q=CV

thus charge increases

potential energy=1/2CV^2

potential energy also increases

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