Question

(a) Derive an expression for the energy stored in a parallel plate capacitor. Hence obtain the expression for the energy density of the electric field.
(b) A fully charged parallel plate capacitor is connected across an uncharged identical capacitor. Show that the energy stored in the combination is less than that stored initially in the single capacitor.

Answer 1

Hint: The main use of the capacitor is to store electrical energy. A charged capacitor stores electrical energy in the electric field between its plates.

 Complete step by step answer:
Consider capacitor of capacitance C being charged from a battery of emf V. Initially, its two plates are uncharged. The positive charge is transferred from the second plate to the first plate in a very small amount of $dq$ till the first plate acquires the final charge +Q and the second plate acquires the charge –Q.
At a certain instant t, during the process of charging, let q be the charge on the plate and V is the potential difference between the plates.
Then $q = CV$
$V = \dfrac{q}{C}$ ……………………… (1)
Where $C$ is the capacitance of the capacitor.
The amount of work done in transferring a small amount of charge $dq$ against the potential difference is,
$dw = Vdq$
Substitute equation (1) in above equation,
$dw = \dfrac{q}{C}dq$ ………………… (2)
The total work done in transferring charge Q is given by,
$W = \int\limits_0^W {dw} $
$W = \int\limits_0^W {\dfrac{q}{C}dq} $
$W = \dfrac{1}{C}\int\limits_0^W {qdq} $
Use this formula for integration,$\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} $
After integration we get,
$W = \dfrac{1}{C}\left[ {\dfrac{{{q^2}}}{2}} \right]_0^Q$
This is the amount of work done is stored as electrostatic potential energy U in the capacitor.
Thus energy stored in the capacitor is $U = \dfrac{1}{C}\left[ {\dfrac{{{Q^2}}}{2}} \right]$
$ = \dfrac{{{Q^2}}}{{2C}}$
We have $Q = CV$ substitute in above equation we get
$U = \dfrac{1}{2}C{V^2}$ These are the different kinds of equations of energy stored in a capacitor.
Energy density: energy stored per unit volume of the space between the plates is called energy density.
consider the separation $d$ of a parallel plate capacitor. Let us have $U = \dfrac{1}{2}C{V^2}$be the energy stored in the space of volume V=$A \times d$ between the plates.
Hence energy density, $u = \dfrac{{energy}}{{volume}}$
$ = \dfrac{{\dfrac{1}{2}C{V^2}}}{{Ad}}$……………………( 3)
But the capacitance $C = \dfrac{{{\varepsilon _0}A}}{d}$ and potential difference $V = Ed$
Substituting in equation (3)
$u = \dfrac{1}{2}\dfrac{{{\varepsilon _0}A}}{d} \times \dfrac{{{E^2}{d^2}}}{{Ad}}$
$u = \dfrac{1}{2}{\varepsilon _0}{E^2}$
This is a general expression which is applicable for electric fields due to any configuration of charges.
Energy stored before connection,
${U_1} = \dfrac{{{Q^2}}}{{2C}}$
${U_2} = 2\dfrac{{{{\left( {\dfrac{Q}{2}} \right)}^2}}}{{2C}}$
Energy stored after connection,
$ = \dfrac{{{Q^2}}}{{4C}}$
${U_2} < {U_1}$
From this we can say, storage of energy in the combination is less than that of the single capacitor.

Note:
Charging a capacitor means transferring electrons from one plate to another.
The capacitance of a capacitor depends on (i) the shape of its plates, (ii) separation between plates, and (iii) the nature of dielectric material between the plates.