A 13.0 μFμF capacitor is connected to a power supply that keeps a constant potential difference of 24.0 VV 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.

Part A

How much energy is stored in the capacitor before the dielectric is inserted?

Part B

How much energy is stored in the capacitor after the dielectric is inserted?

Part C

By how much did the energy change during the insertion?

a) 3.744×10^-3Joules

b) 5.265 × 10^-2Joules

c) 0.0489Joules

Explanation:

Capacitors are devices used for storing electric charge.

Energy stored in a capacitor is expressed as Eo = 1/2CVo² where:

C is the capacitance of the capacitor

V is the potential difference across the capacitor

a) Energy stored in the capacitor before the dielectric is inserted can be gotten using the formula above.

Given C = 13×10^-6F

Vo = 24V

Energy stored in the capacitor = 1/2 × 13×10^-6 × 24²

Eo = 3744×10^-6Joules

Eo = 3.744×10^-3Joules

b) Energy is stored in the capacitor after the dielectric is inserted can be calculated using:

E = 1/2CV²

New potential difference V = kVo where k is dielectric constant

Vo is the potential difference across the capacitor with out the dielectric

V = 3.75×24

V = 90Volts

E = 1/2×13×10^-6×90²

E = 52,650×10^-6

E = 5.265 × 10^-2Joules

c) Energy change during insertion

= E-Eo

= 5.265 × 10^-2 - 3.744×10^-3

= 0.05265-0.003744

= 0.0489Joules