Question

(a) Derive the 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

Consider a capacitor of capacitance C. Initial charge on capacitor is zero. Initial potential difference between capacitor plates is zero. Let a charge Q be given to it in small steps. When charge is given to capacitor, the potential difference between its plates increases. Let at any instant when charge on capacitor be q, the potential difference between its plates $V=\frac{q}{C}$.



Now work done in giving an additional infinitesimal charge dq to capacitor.

$dW=Vdq=\frac{q}{C}dq$

$W={\int }_0^{Q}Vdq={\int }_0^Q\frac{q}{C}dq$

$=\frac{1}{C}{\left[\frac{q^2}{2}\right]}_0^Q=\frac{1}{C}(\frac{Q^2}{2}-\frac{O}{2})=\frac{Q^2}{2C}$

If V is the final potential difference between capacitor plates, then Q=CV

$\therefore W=\frac{(CV{)}^2}{2C}=\frac{1}{2}C{V}^2=\frac{1}{2}QV$

This work is stored as electrostatic potential energy of capacitor i.e.,

Electrostatic potential energy,

$U=\frac{Q^2}{2C}=\frac{1}{2}C{V}^2=\frac{1}{2}QV$

Energy density : Consider a parallel plate capacitor consisting of plates, each of area A, separated by a distance d. If space between the plates is filled with a medium of dielectric constant K, then

Capacitance of capacitor, $C=\frac{K{\epsilon }_{0}A}{d}$

If $\sigma$ is the surface charge density of plates, then electric field strength between the plates

$E=\frac{\sigma }{K{\epsilon }_0}\Rightarrow \sigma =K{\epsilon }_{0}E$

Charge on each plate of capacitor, $Q=\sigma A=K{\epsilon }_{0}EA$

$\therefore$ Energy stored by capacitor,

$U=\frac{Q^2}{2C}=\frac{(K{\epsilon }_{0}EA{)}^2}{2\left(K{\epsilon }_{0}A/d\right)}$

$=\frac{1}{2}K{\epsilon }_0{E}^{2}Ad$

But Ad= Volume of space between capacitor plates

$\therefore$ Energy stored, $U=\frac{1}{2}K{\epsilon }_0{E}^{2}Ad$

Electrostatic energy stored per unit volume,

$u_e=\frac{U}{Ad}=\frac{1}{2}K{\epsilon }_0{E}^2$

This is expression for electrostatic energy density in medium of dielectric constant K.

(b) Initially, if we consider a charged capacitor, then its charge would be

Q=CV


and energy stored is

$U_1=\frac{1}{2}C{V}^2$ ...(i)

Then, this charged capacitor is connected to uncharged capacitor.

Let the common potential be V. The charge flows from first capacitor to the second capacitor untill both the capacitors attain common potential.


$\Rightarrow Q_1=C{V}_1$ and $Q_2=C{V}_2$

Applying conservation of charge,

$Q=Q_1+Q_2\Rightarrow CV=C{V}_1+C{V}_2$

$\Rightarrow V=V_1+V_2\Rightarrow V_1=\frac{V}{2}$

Total energy stored,

$U_2=\frac{1}{2}C{V}_1^2+\frac{1}{2}C{V}_2^2$

$U_2=\frac{1}{2}C{\left(\frac{V}{2}\right)}^2+\frac{1}{2}C{\left(\frac{V}{2}\right)}^2$

$\Rightarrow U_2=\frac{1}{4}C{V}^2$ ...(ii)

From Equations (i) and (ii), we get

$U_2<U_1$

$\Rightarrow$ Energy stored in the combination is less than that stored initially in single capacitor.