Question

Learning Goal:

Answer 1

A) $U_0 = \frac{\epsilon_0 A V^2}{2d}$

B) $U_1=\frac{3\epsilon_0 A V^2}{2d}$

C) $U_2=\frac{k\epsilon_0 A V^2}{2d}$

Explanation:

A)

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

$C=\frac{\epsilon_0 A}{d}$ (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

$U_0 = \frac{1}{2}CV^2$ (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:

$U_0 = \frac{\epsilon_0 A V^2}{2d}$

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:

$Q=CV=\frac{\epsilon_0 A V}{d}$ (1)

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

$U=\frac{Q^2}{2C}$ (3)

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

$C=\frac{\epsilon_0 A}{3d}$ (2)

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

$U_1 = \frac{(\frac{\epsilon_0 A V}{d})^2}{2(\frac{\epsilon_0 A}{3d})}=\frac{3\epsilon_0 A V^2}{2d}$

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

$C=\frac{k\epsilon_0 A}{d}$

where

k is the dielectric constant of the dielectric material

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

$U=\frac{1}{2}CV^2$

And substituting C and V, we find

$U_2=\frac{k\epsilon_0 A V^2}{2d}$