Question

A parallel plate capacitor, each of plate area A and separation d between the two plates, is charged with charges +Q and -Q on the plates. Deduce the expression for the energy stored in the capacitor.

Answer 1

Capacitor is a device used to store electrical energy in form of electric charges. The capacitance of a parallel plate capacitor depends upon the area, distance between the two plates and permittivity of the medium. Then substitute this in the final energy expression found with the help of voltage across the plates.

Complete step by step answer:
Capacitance is defined as the measure to hold electric charges. Mathematically, it is defined as the ratio of the charge given to rise the potential of the conductor i.e.
$C = \dfrac{Q}{V}$
C is the capacitance and is dependent on the shape, size and presence of conductors or dielectrics inside as well as in nearby space. For a parallel plate capacitor, each plate act as charged conducting sheet for which the electric field is defined as:
$E = \dfrac{\sigma }{{{\varepsilon _o}}}$
Where, $\sigma = Q/A$ and the potential in terms of electric field is defined as V=Ed so, on substituting, we have,

$$V = \dfrac{\sigma }{{{\varepsilon _o}}}d = \dfrac{{Qd}}{{{\varepsilon _o}A}} \\\$$

Therefore,

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

For the energy stored in capacitor,
It is defined as the work done in charging the capacitor from potential zero to potential V, so, in electrostatics the work done is defined as the energy spent in moving charge in potential field, so

$$W = VQ \\\ dW = VdQ \\\$$

If a battery/cell of potential V is connected to a capacitor then the potential is constant across the capacitor but the charge is increasing from zero to say Q, so,

$$dW = VdQ = \dfrac{Q}{C}dQ \\\ W = \int\limits_0^Q {\dfrac{Q}{C}dQ} = \left[ {\dfrac{{{Q^2}}}{{2C}}} \right]_0^Q = \dfrac{1}{2}\dfrac{{{Q^2}}}{C} \\\ or \\\ W = \dfrac{1}{2}C{V^2} \\\$$

Therefore, the energy stored in the capacitor is defined as:

$$U = \dfrac{1}{2}C{V^2} = \dfrac{{{Q^2}}}{{2C}}joule \\\ U = \dfrac{1}{2}\dfrac{{{\varepsilon _o}A}}{d}{V^2} = \dfrac{1}{2}\dfrac{d}{{{\varepsilon _o}A}}{Q^2} \\\$$

Note: For a capacitor the capacitance remains constant unless the shape or size or any conductor, dielectrics are not introduced. Therefore, the energy depends upon the capacitance and the potential or charge stored. When a dielectric is introduced between the plates then the capacitance becomes:
$C = \dfrac{{K{\varepsilon _o}A}}{d}$