Question

The energy stored per unit volume between two plates of a parallel

plate capacitor will be

a) (1/2ϵ o )E

b) (½ ϵ o2 )E

c) (½ ϵ o2)E2

d) (½ ϵ o)E2​

Answer 1

20

Explanation:

shared pair of electrons so there are 2electrons in each

Answer 2

Answer:

The energy stored per unit volume between two plates of the parallel

plate capacitor will be $\frac{1}{2} \epsilon_oE^2$ i.e.option(d).

Explanation:

The energy stored in the capacitor is calculated as follows,

$K=\frac{1}{2} CV^2$          (1)

Where,

K=energy stored in the capacitor

C=capacitance of the capacitor

V=potential difference between the plates of the capacitor

Also, we know,

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

∈₀=permitivity of free space

A=area between the plates of the capacitor

d=distance between the plates of the capacitor

By putting equation (2) in equation (1) we get;

$K=\frac{1}{2} \times \frac{\epsilon_o A}{d}\times V^2$         (3)

And,

The volume of the plates=Ad        (4)

And the electric field in terms of the potential is given as,

$E=\frac{V}{d}$          (5)

So, the energy per unit volume is given as,

$\frac{K}{Ad} =\frac{1}{2} \times \frac{\epsilon_o A}{d}\times V^2\times \frac{1}{Ad}$

$\frac{K}{Ad}=\frac{1}{2}\times \epsilon_o\times \frac{V^2}{d^2}$         (6)

By putting equation (5) in equation (6) we get;

$\frac{K}{Ad}=\frac{1}{2} \epsilon_oE^2$

Hence, the energy stored per unit volume between two plates of the parallel

plate capacitor will be $\frac{1}{2} \epsilon_oE^2$ i.e.option(d).