Physics, 25.03.2020 06:07 Animallover100

Energy of a capacitor. [20] A parallel-plate capacitor has area A and plate separation d. (a) The capacitor is connected to a battery with voltage (EMF) E. What is the potential energy of the capacitor? [6] (b) The capacitor remains connected to the battery, and a polymer slab with thickness d and relative permittivity κ is inserted into the gap. What is the new potential energy of the capacitor? [6] (c) The dielectric is removed, then the battery is disconnected, and the dielectric is reinserted while the battery remains disconnected. What is the internal energy of the capacitor after this process? [6] (d) What is the change of the potential energy of the battery during the process described in part (b)? Hint: work is not zero. [2]

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Answer from: johndoesnutz4690

a) $U=\frac{\epsilon_0 A E^2}{2d}$

b) $U=\frac{\epsilon_0 k AE^2}{2d}$

c) $U=\frac{\epsilon_0 AE^2}{2kd}$

d) $\frac{\epsilon_0 AE^2}{2d}(1-k)$

Explanation:

The electric potential energy stored in a capacitor is given by:

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

where

C is the capacitance

V is the potential difference across the capacitor

For a parallel-plate capacitor, the capacitance is

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

where

$\epsilon_0$ is the vacuum permittivity

A is the area of the plates

d is the separation between the plates

Moreover in this problem, the potential difference across the capacitor is equal to the EMF of the battery:

V=E

Substituting the last two expressions into eq.(1), we find the potential energy in the capacitor:

$U=\frac{\epsilon_0 A E^2}{2d}$

In this second part, a polymer slab with thickness d (the same as the separation between the plates) and relative permittivity k is inserted into the capacitor.

When a dielectric is inserted into a capacitor, its capacitance increases due to the polarization effect on the molecules of the dielectric, according to the equation

C'=kC

where

C' is the new capacitance

k is the relative permittivity

C is the original capacitance

In this case, the dielectric is inserted while the battery is still connected: this means that the potential difference across the capacitor remains the same, only the capacitance changes.

If we look at eq.(1), we see that the potential energy is directly proportional to the capacitance of the capacitor. Therefore, since the capacitance has increased by a factor k, the potential energy will also increase by a factor k, therefore:

$U'=\frac{1}{2}C'V^2=\frac{\epsilon_0 k AE^2}{2d}$

In this case, the dielectric is removed, the battery disconnected, and the dielectric is inserted again keeping the battery disconnected.

This means that the potential difference V does not remain the same across the capacitor: however, the charge stored remains the same (since the capacitor is disconnected from everything, it cannot discharge).

The initial charge stored in the capacitor is

Q=CV

where

$C=\frac{\epsilon_0 A}{d}$ is the initial capacitance

V=E is the initial potential difference

$Q=\frac{\epsilon_0 A E}{d}$

The charge remains constant, while the capacitance changes when the dielectric is inserted, and it becomes

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

where k is the relative permittivity.

To find the new energy stored in the capacitor, we use the other formula:

$U'=\frac{Q^2}{2C'}$

which is equivalent to eq.(1). Substituting everything, we find:

$U'=\frac{(\frac{\epsilon_0 AE}{d})^2}{2(\frac{k\epsilon_0 A}{d})}=\frac{\epsilon_0 AE^2}{2kd}$

Here we want to find the change in the potential energy of the battery during the process described in part b).

Due to the law of conservation of energy, the total energy of the battery + the capacitor must be constant. This means that

$\Delta U_C + \Delta U_B =0$ (2)

where

$\Delta U_C$ is the change in potential energy of the capacitor

$\Delta U_B$ is the change in potential energy of the battery

From part a) and b), we can find the change in potential energy of the capacitor:

$\Delta U_C=U'-U=\frac{\epsilon_0 kAE^2}{2d}-\frac{\epsilon_0 AE^2}{2d}=\frac{\epsilon_0 AE^2}{2d}(k-1)$

And so according to eq.(2), the change in the potential energy of the battery is

$\Delta U_B = -\Delta U_C=-\frac{\epsilon_0 AE^2}{2d}(k-1)=\frac{\epsilon_0 AE^2}{2d}(1-k)$

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Answer from: Quest

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Energy of a capacitor. [20] A parallel-plate capacitor has area A and plate separation d. (a) The ca...

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