Two spherical conductors of capacities $3 \mu \mathrm{F}$ and $2 \mu \mathrm{F}$ are charged to same…
Two spherical conductors of capacities $3 \mu \mathrm{F}$ and $2 \mu \mathrm{F}$ are charged to same potential having radii $3 \mathrm{~cm}$ and $2 \mathrm{~cm}$ respectively. If ' $\sigma_1$ ' and ' $\sigma_2$ ' represent surface density of charge on respective conductors then $\frac{\sigma_1}{\sigma_2}$ is
$\frac{1}{3}$
$\frac{1}{2}$
$\frac{2}{3}$
$\frac{3}{4}$
Solution
We know, $\mathrm{C}=\frac{\mathrm{Q}}{\mathrm{V}}$
As both the charged spheres are at the same potential, the charge on both spheres is
$\begin{aligned}
& Q_1=C_1 \mathrm{~V} \\
& Q_2=C_2 \mathrm{~V}
\end{aligned}$
The charge densities of both spheres are
$\sigma_1=\frac{\mathrm{Q}_1}{\mathrm{~A}_1}=\frac{\mathrm{C}_1 \mathrm{~V}}{4 \pi \mathrm{r}_1^2}$
Similiarly,
$\sigma_2=\frac{\mathrm{Q}_2}{\mathrm{~A}_2}=\frac{\mathrm{C}_2 \mathrm{~V}}{4 \pi \mathrm{r}_2^2}$
Taking the ratios,
$\begin{aligned}
\frac{\sigma_2}{\sigma_1} & =\frac{\mathrm{C}_2 \mathrm{r}_1^2}{\mathrm{C}_1 \mathrm{r}_2^2} \\
\frac{\sigma_2}{\sigma_1} & =\frac{2 \times 10^{-6} \times(0.03)^2}{3 \times 10^{-6} \times(0.02)^2}=\frac{3}{2} \\
\therefore \quad \frac{\sigma_1}{\sigma_2} & =\frac{2}{3}
\end{aligned}$