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

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

Answer 2

After placing the dielectrics between the plates of the capacitor, the energy stored in the capacitor becomes 0.5 J.

What is capacitance?

Capacitance is a term used to define the amount of energy stored in the form of an electric charge in an electric device.

Given data:

The capacitance of parallel-plate capacitor is, C = 10 mF = 0.01 F.

The potential of power supply is, V' = 20 V.

The dielectric constant of the material is, ∈ = 4.0.

Let us first calculate the charge stored in the capacitor as,

$q = CV'\\\\q = 0.01 \times 20\\\\q = 0.2 \;\rm C$

After placing the dielectric, the capacitance of the capacitor is,

$C' = \epsilon \times C$

Solving as,

$C' = 4.0 \times 0.01\\\\C' = 0.04 \;\rm F$

Now, the expression for the energy stored in the capacitor is,

$U'=\dfrac{q^{2}}{2C'}$

Solving as,

$U'= \dfrac{0.2^{2}}{2 \times 0.04}\\\\U'= 0.5 \;\rm J$

Thus, we can conclude that the energy stored in the capacitor is of 0.5 J.

Learn more about the capacitance here:

https://brainly.com/question/12356566