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

Answer 1

1. increase the area of the plates

4. decrease the separation between the plates

5. insert a dielectric between the plates

Step-by-step explanation:

The energy of a capacitor is given by:

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

where
`C` is the capacitance and
`V` is the voltage or potential difference between the plates. With a constant potential difference, the energy is proportional to the capacitance i.e. the higher the capacitance, the higher the energy and vice-versa.

Now the capacitance has a geometric formula of:

`C = \epsilon(A)/(d)`

where

`\epsilon` is the permittivity of the material between the plates of the capacitor,
`A` is the area of the plates and
`d` is the separation distance between the plates.

From the relation, it is seen that the capacitance is directly proportional to the permittivity and the area but inversely proportional to the separation. In order to increase the capacitance, the permittivity or/and the area of the plates should be increased while the separation should be decreased.