Question

(a) Derive the expression for the capacitance of a parallel plate capacitor having plate area A and plate separation d.
(b) Two charged spherical conductors of radii $R_1$ and $R_2$ when connected by a conducting wire acquire charge $q_1$ and $q_2$ respectively. Find the ratio of their surface charge densities in terms of their radii.

Answer 1

(a)

Electric field on the inside of the parallel plate capacitor due to one of the plates is given by:

$E_1=\frac{\sigma }{2{\epsilon }_o}=\frac{Q}{2A{\epsilon }_o}$

Total electric field due to the two plates is given by:

$E=2E_1=\frac{Q}{A{\epsilon }_o}$

Potential difference between the plates is given by:

$V=Ed$

$V=\frac{Qd}{A{\epsilon }_o}$

By definition,

$C=\frac{Q}{V}=\frac{A{\epsilon }_o}{d}$

(b)

After connection, both spheres are at the same potential.

$V_1=V_2$

$\frac{K{q}_1}{R_1}=\frac{K{q}_2}{R_2}$

$\frac{q_1}{q_2}=\frac{R_1}{R_2}$

Surface charge density for a spherical conductor is given by:

$\sigma =\frac{Q}{S}$

Hence, ${\sigma }_1=\frac{q_1}{4\pi R_1^2}$

${\sigma }_2=\frac{q_2}{4\pi R_2^2}$

$\frac{{\sigma }_1}{{\sigma }_2}=\frac{q_1{R}_2^2}{q_2{R}_1^2}=\frac{R_2}{R_1}$