Question

A 13.5 μF capacitor is connected to a power supply that keeps a constant potential difference of 22.0 V across the plates. A piece of material having a dielectric constant of 3.75 is placed between the plates, completely filling the space between them.

a. How much energy is stored in the capacitor before the dielectric is inserted?
b. How much energy is stored in the capacitor after the dielectric is inserted?
c. By how much did the energy change during the insertion?

Answer 1

Step-by-step explanation:

A. The amount of energy stored before the dielectric is added;

Using formula;

W = 1/2CV^2

Where, W= energy

C = capacitance of a capacitor

V= Potential difference

W = 1/2 CV^2

W = 1/2 * 13.5*10^-6F * 22.0^2 V

W = 1/2 * 13.5*10^-6 * 484

W = 1/2 * 6534*10^-6

W = 3267*10^-6Joules

The energy stored before inserting the dielectric material is therefore 3.267*10^-3 Joules.

B. After the dielectric is added, how much energy is stored?

Dielectric constant (k) = 3.75

Note: when the dielectric material is added, the capacitance increased to;

C new = k C initial

C new = 3.75 * 13.5*10^-6F

C new = 5.0625*10^-5F

Also, the potential difference will change

V = 1/k * V initial

V = 1/3.75 * 22 V

V =5.867 V

So therefore, the energy stored when the dielectric material is added will be;

W = 1/2 CV^2

W = 1/2 * 5.0625*10^-5 * 5.867^2

W = 1/2 *5.0625*10^-5 * 34.421

W = 8.71299*10^-4 Joules

W = ~ 8.71*10^-4 Joules

C. How much is the energy change?

Energy change = Energy before insertion - Energy after insertion

= 3.267*10^-3 J - 8.713*10^-4 J

= 3.267*10^-3 - 0.8713*10^-3

= 2.3957*10^-3

Energy change is approximately 2.4*10^-3 J