Question

Learning Goal:

To understand that the charge stored by capacitors represents energy; to be able to calculate the stored energy and its changes under different circumstances.

An air-filled parallel-plate capacitor has plate area A and plate separation d. The capacitor is connected to a battery that creates a constant voltage V.

Part A

Find the energy U0 stored in the capacitor.

Express your answer in terms of A, d, V, and ?0. Remember to enter ?0 as epsilon_0.

Part B

The capacitor is now disconnected from the battery, and the plates of the capacitor are then slowly pulled apart until the separation reaches 3d. Find the new energy U1 of the capacitor after this process.

Express your answer in terms of A, d, V, and ?0.

Part C

The capacitor is now reconnected to the battery, and the plate separation is restored to d. A dielectric plate is slowly moved into the capacitor until the entire space between the plates is filled. Find the energy U2 of the dielectric-filled capacitor. The capacitor remains connected to the battery. The dielectric constant is K.

Express your answer in terms of A, d, V, K, and ?0.

Answer 1

A)
`U_0 = (\epsilon_0 A V^2)/(2d)`

B)
`U_1=(3\epsilon_0 A V^2)/(2d)`

C)
`U_2=(k\epsilon_0 A V^2)/(2d)`

Step-by-step explanation:

A)

First of all, the capacitance of a parallel-plate capacitor filled with air is given by

`C=(\epsilon_0 A)/(d)` (1)

where

`\epsilon_0` is the vacuum permittivity

A is the area of the plates

d is the separation between the plates

The energy stored by a capacitor is given by

`U_0 = (1)/(2)CV^2` (2)

where

C is the capacitance

V is the potential difference across the plates

Substituting (1) into (2), we find an expression of the tenergy stored:

`U_0 = (\epsilon_0 A V^2)/(2d)`

B)

In this part of the problem, the capacitor is disconnected from the battery.

This means that now the charge on the capacitor remains constant. The charge can be written as

`Q=CV`

Since the charge is the same as in part A), we can write it explicitely:

`Q=CV=(\epsilon_0 A V)/(d)` (1)

We can write the energy stored in the capacitor using another equation:

`U=(Q^2)/(2C)` (3)

In this case, the distance between the plates is increased to 3d, so the new capacitance is

`C=(\epsilon_0 A)/(3d)` (2)

Substituting (1) and (2) into (3), we find the new energy stored:

`U_1 = (((\epsilon_0 A V)/(d))^2)/(2((\epsilon_0 A)/(3d)))=(3\epsilon_0 A V^2)/(2d)`

3)

In this case, the capacitor is reconnected to the battery, so the potential difference is now equal to the initial potiential difference V.

In this case, however, a dielectric plate is moved inside the space between the plates. Therefore, the capacitance becomes

`C=(k\epsilon_0 A)/(d)`

where

k is the dielectric constant of the dielectric material

To calculate the energy stored, we can use again the original formula

`U=(1)/(2)CV^2`

And substituting C and V, we find

`U_2=(k\epsilon_0 A V^2)/(2d)`