Question

A 13.5 μF capacitor is connected to a power supply that keeps a constant potential difference of 24.0 V across the plates. A piece of material having a dielectric constant of 3.55 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? Did it increase or decrease?
D) Explain why inserting the dielectric (or equivalently exchanging air with the material) causes a change in the stored energy of the capacitor?

Answer 1

Answer:

Explanation:

Capacitance of the capacitor = 13.5μF

Voltage across plate is 24V

Dielectric constant k=3.55.

a. Energy in capacitor is given by

E=1/2CV^2

We want to calculate energy without the dielectric substance

Given that C=13.5 μF and V=24V

The capacitance give is with dielectric so we need to remove it

C=kCo

Co=C/k

Then the Co=13.5μF/3.55

Co=3.803μF

Then

E=(1/2)×3.803×10^-6×24^2

E=1.1×10^-3J

E=1.1mJ

b. Energy in capacitor is given by

E=1/2CV^2

The capacitance given is with a dielectric, so we are going to apply it direct.

Given that C=13.5 μF and V=24V

Then

E=(1/2)×13.5×10^-6×24^2

E=3.89×10^-3J

E=3.9mJ

c. The energy without dielectric is 1.1mJ and the energy with dielectric is 3.9mJ

The energy increase when the dielectric material is added

d. Dielectrics in capacitors serve three purposes: to keep the conducting plates from coming in contact, allowing for smaller plate separations and therefore higher capacitances;

Therefore, Since dielectric allow higher capacitance, and energy of a capacitor is directly proportional to the capacitance, then the higher the capacitance the higher the energy.