Question

A parallel plate capacitor is charged, such that electric field between the plates of capacitor is E. The energy density of the capacitor is​

Answer 1

Answer:

The energy density is given as,

u

E

=

2

1

ε

0

E

2

1.8×10

−9

=

2

1

×9×10

−12

E

2

E

2

=400

=20NC

−1

Thus, the value of the electric field is 20NC

−1

.

Answer 2

Answer:

The energy density is  $u = {\dfrac{\varepsilon E^{2}}{2}}$

Explanation:

Given a parallel plate capacitor is charged, and the electric field between the plates of capacitor is E.

• For a parallel-plate capacitor, the maximum amount of charge that can be acquired is proportional to the voltage.

        $Q\propto V\implies Q=CV$

        where C, the proportionality constant is called capacitance.

• The capacitance of the capacitor depends on its geometry, given by

        $C = \dfrac{\varepsilon A}{d}$

        where $A$ is the area of a plate, $d$ is the distance between the plates,   and $\varepsilon$ is the electric permittivity.

• Capacitors store electric energy when they are charged. And the energy stored is the work done to move a charge Q through a potential difference V.
• The electric potential energy stored in a parallel plate capacitor is given by the charge times the average potential difference.

        $\therefore U = \frac{1}{2} QV= \frac{1}{2} (CV)V=\frac{1}{2} \dfrac{\varepsilon A}{d}V^{2}$

• For a parallel-plate capacitor, the voltage is proportional to the electric field, i.e., $V=Ed$

        $\therefore U = \frac{1}{2} QV= \dfrac{\varepsilon A}{2d} (Ed)^{2}= \dfrac{\varepsilon AE^{2}d}{2}$

• The energy stored in a capacitor also depends on the capacitor's geometry given by A and d.  
• The energy density u in a capacitor is defined as the energy stored per unit volume is given by

       $u = \dfrac{U}{volume} =\dfrac{\dfrac{\varepsilon AE^{2}d}{2}}{Ad} ={\dfrac{\varepsilon E^{2}}{2}}$

Hence, the energy density is given by $u = {\dfrac{\varepsilon E^{2}}{2}}$ and depends only on the electric field and the permittivity medium.