Question

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

Answer 1

A) Energy density of a parallel plate capacitor is defined as energy stored in the electric fields of the capacitor per unit volume.
B) To derive an expression for the energy stored in a parallel plate capacitor:
Work done in depositing $dq$ amount of charge over the plates at potential difference $V$ is given by-
$dW=Vdq$
Now, $q=CV$ or $V=\frac{q}{C}$
$dW=\frac{q}{C}dq$
Total work done in depositing $q$ charge over the plates is
$W=\int dW={\int }_0^q\frac{q}{C}dq=\frac{q^2}{2C}$
This work done is stored as the energy of the capacitor
$\therefore E=\frac{q^2}{2C}=\frac{1}{2}qV=\frac{1}{2}C{V}^2$
C) For a parallel plate capacitor:
$C=\frac{{ϵ}_{0}A}{d}$
$\therefore A=\frac{Cd}{{ϵ}_0}=\frac{0.1\times {10}^{-6}\times 0.1\times {10}^{-3}}{8.854\times {10}^{-12}}=1.13m^2$