A parallel-plate capacitor has a capacitance of 10 mF and charged with a 20-V power supply.The power supply is then removed and a dielectric material of dielectric constant 4.0 is used to fill the space between the plates.How much energy is now stored by the capacitor?
A) 250 mJ
B) 125 mJ
C) 500 mJ
D) 62.5 mJ
E) 1200 mJ

Question

Quizplus · Accepted Answer

The energy stored by a capacitor is given by the formula:

U = 1/2 * C * V^2

where U is the energy, C is the capacitance, and V is the voltage.

Initially, the capacitor is charged to a voltage of 20V, so the initial energy stored is:

U1 = 1/2 * 10 mF * (20V)^2 = 2 mJ

When the dielectric material is inserted, the capacitance increases by a factor of 4.0, so the new capacitance is:

C' = 4.0 * C = 40 mF

Since the voltage across the capacitor is unchanged, the final energy stored is:

U2 = 1/2 * 40 mF * (20V)^2 = 8 mJ

The change in energy is therefore:

ΔU = U2 - U1 = 6 mJ

So the answer is:

C) 500 mJ

A Parallel-Plate Capacitor Has a Capacitance of 10 MF and Charged