Question

A. What is meant by the energy density of a parallel plate capacitor?
B. Derive an expression for the energy stored in parallel plate capacitor
C. What is the area of plates $0.1\mu F$ parallel plate air capacitor, given that the separation between plates is 0.1mm?

Answer 1

Hint: Capacitor is a device which is used to store the charge. The charge stored in the capacitor depends on the potential difference applied across them and also depends on the capacitance of the capacitor. We will use the relation between them to derive the required expression.
Formula used:
$q = cV$

Complete answer:
Energy density of a parallel plate capacitor is nothing but the ratio of amount of energy stored in the capacitor when charge is present on the plates of the capacitor to the volume of the capacitor.
The charge stored on the plates of the capacitor is given by $q = cV$. When the battery of emf V is connected to the capacitor, work will be done by the battery to charge the capacitor. The amount of work done will be converted to the energy stored in the capacitor.
Let us assume dq amount of charge is accumulated on the plates of the capacitor. The work done by battery in doing so will be $dW = dqV$
We will integrate that expression from initial state to the state where potential difference across the capacitor becomes ‘V’ to get the total energy stored
$\eqalign{
  & dW = dqV \cr
  & \Rightarrow q = cV \cr
  & \Rightarrow dW = \left( {cV} \right)dV \cr
  & \Rightarrow \int\limits_0^W {dW} = \int\limits_0^V {\left( {cV} \right)dV} \cr
  & \Rightarrow W = \dfrac{1}{2}c{V^2} \cr
  & \Rightarrow U = \dfrac{1}{2}c{V^2} \cr} $
So the energy stored in the capacitor will be $U = \dfrac{1}{2}c{V^2}$
Capacitance is given by
$c = \dfrac{{{\varepsilon _0}A}}{d}$
Where A is the area of plates and ‘d’ is the distance between plates.
$\eqalign{
  & c = \dfrac{{{\varepsilon _0}A}}{d} \cr
  & \Rightarrow A = \dfrac{{dc}}{{{\varepsilon _0}}} \cr
  & \Rightarrow A = \dfrac{{\left( {0.1 \times {{10}^{ - 3}}} \right)\left( {0.1 \times {{10}^{ - 6}}} \right)}}{{8.85 \times {{10}^{ - 12}}}} \cr
  & \Rightarrow A = 1.1299{m^2} \cr} $
Hence the area of plates will be 1.1299 square meter

Note:
There is a formula for the energy density of the capacitor too. We get it by substituting the potential difference with the product of electric field and distance between the plates. We divide the energy stored in the capacitor with the volume to get it. It will be $\dfrac{1}{2}{\varepsilon _0}{E^2}$ where E is the electric field.