A parallel-plate capacitor carries a constant charge Q (i.e., +Q on one plate, a
ID: 2187390 • Letter: 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