Question

A 12.0 μF capacitor is connected to a power supply that keeps a constant potential difference of 26.0 V across the plates. A piece of material having a dielectric constant of 3.80 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 or1 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

Step-by-step explanation:

Capacitance, C = 12 micro farad

voltage, V = 26 V

dielectric constant, K = 3.8

(a) The formula for the energy stored is

U = 1/2 CV²

U = 0.5 x 12 x 10^-6 x 26 x 26

U = 0.00406 J

(b) U' = 1/2 C'V²

U' = 0.5 x K x C x V² = K U

U' = 3.8 x 4.056 x 10^-3

U' = 0.0154 J

(c) The energy increases. As the value of dielectric constant increases so the energy increases.

Answer 2

0.004056 J

0.0154128 J

Increase

Step-by-step explanation:

C = Capacitance = 12 μF

V = Potential difference = 26 V

K = Dielectric constant = 3.8

New capacitance is

`C'=KC\\\Rightarrow C'=3.8* 12* 10^(-6)\\\Rightarrow C'=0.0000456\ F`

Energy in the capacitor

`U=(1)/(2)CV^2\\\Rightarrow U=(1)/(2)* 12* 10^(-6)* 26^2\\\Rightarrow U=0.004056\ J`

Before insertion energy 0.004056 J

After insertion

`U'=(1)/(2)C'V^2\\\Rightarrow U'=(1)/(2)* 0.0000456* 26^2\\\Rightarrow U'=0.0154128\ J`

After insertion energy 0.0154128 J

Change in energy

`\Delta E=U'-U\\\Rightarrow \Delta E=0.0154128-0.004056\\\Rightarrow \Delta E=0.0113568\ J`

There is an increase in energy.

The dielectric constant of air is about 1.00059 here the dielectric constant is 3.8 which is higher so the energy increases.