Question

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?

Answer 1

Answer:

0.5 J

Explanation:

For this capacitor we have:

$C=10 mF = 0.01 F$ is the capacitance

V = 20 V is the potential difference

So the charge stored in the capacitor is

$Q=CV=(0.01 F)(20 V)=0.2 C$

Later, the power supply is removed, so the charge on the capacitor will remain the same. A dielectric of dielectric constant

k = 4.0

is inserted in the gap between the plates. The capacitance of the capacitor change as follows:

C' = k C = (4.0)(0.01 F) = 0.04 F

The energy stored in the capacitor is given by

$U'=\frac{1}{2}\frac{Q^2}{C'}$

and using Q = 0.2 C, we find

$U'=\frac{1}{2}\frac{(0.2 C)^2}{(0.04 F)}=0.5 J$