Question

Which two, or more, of the following actions would increase the energy stored in a parallel plate capacitor when a constant potential difference is applied across the plates?
1. increase the area of the plates
2. decrease the area of the plates
3. increase the separation between the plates
4. decrease the separation between the plates
5. insert a dielectric between the plates


Choices
A) 2,4
B) 1.3
C) 1,4, 5
D) 2, 3, 5

Answer 1

1. increase the area of the plates

4. decrease the separation between the plates

5. insert a dielectric between the plates

Explanation:

The energy stored in a capacitor is given by

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

where

C is the capacitance

V is the potential difference across the capacitor

For a parallel-plate capacitor, the capacitance is given by

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

where

k is the dielectric constant of the material

$\epsilon_0$ is the vacuum permittivity

A is the area of the plates of the capacitor

d is the separation between the plates

So we can rewrite the energy stored in the capacitor as

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

Here the potential difference is kept constant, so the energy depends only on the dielectric constant of the medium, the area and on the distance between the plates. In particular:

- The energy is directly proportional to the area, so as the area increases, the energy will increase

- The energy is inversely proportional to the distance, so as the distance decreases, the energy will increase

- The energy increases if the value of k increases (that is, if a dielectric is put between the plates)

It follows that the correct options to increase the energy are:

1. increase the area of the plates

4. decrease the separation between the plates

5. insert a dielectric between the plates

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