Two charged spherical conductors of radii $\mathrm{R}_{1}$ and $R_{2}$ are connected by a wire. Then

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Two charged spherical conductors of radii $\mathrm{R}_{1}$ and $R_{2}$ are connected by a wire. Then the ratio of surface charge densities of the spheres $\sigma_{1} / \sigma_{2}$ is\

Professor PC Thomas Classes · Answer

Potential at the surface of spherical conductor of radius $R$ carrying charge $Q, V_{s}=\frac{Q}{4 \pi \varepsilon_{0} R}$.\ Let $Q_{1}$ and $Q_{2}$ are the charge on two spherical conductor of radii $R_{1}$ and $R_{2}$ respectively. when these two charged spherical conductors connected by a wire, the potential at their surfaces becomes equal.\ $	herefore \mathrm{V}_{\mathrm{S}_{1}}=\mathrm{V}_{\mathrm{S}_{2}}$\ $\frac{\mathrm{Q}_{1}}{4 \pi \varepsilon_{0} \mathrm{R}_{1}}=\frac{\mathrm{Q}_{2}}{4 \pi \varepsilon_{0} \mathrm{R}_{2}}$ $$\begin{equation*} \frac{\mathrm{Q}_{1}}{\mathrm{Q}_{2}}=\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}} 	ag{i} \end{equation*} $$ Surface charge density, $\sigma=\frac{	ext { Charge }}{	ext { Area }}$\ $	herefore \frac{\sigma_{1}}{\sigma_{2}}=\frac{\left(\frac{Q_{1}}{4 \pi R_{1}^{2}}ight)}{\left(\frac{Q_{2}}{4 \pi R_{2}^{2}}ight)}=\frac{Q_{1}}{Q_{2}} 	imes \frac{R_{2}^{2}}{R_{1}^{2}}$\