Two charged spherical conductors of radii R_1 | vedclass

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Question

Potential at the surface of spherical conductor of radius $\mathrm{R}$ carrying charge $\mathrm{Q},$ $\mathrm{V}_{\mathrm{s}}=\frac{\mathrm{Q}}{4 \pi

Vedclass · Accepted Answer

$\frac{{{R_2}}}{{{R_1}}}$

Vedclass · Answer

Potential at the surface of spherical conductor of radius $\mathrm{R}$ carrying charge $\mathrm{Q},$ $\mathrm{V}_{\mathrm{s}}=\frac{\mathrm{Q}}{4 \pi \varepsilon_{0} \mathrm{R}}$

Let $Q_{1}$ and $Q_{2}$ are the charges on two spherical conductors of radii $\mathrm{R}_{1}$ and $\mathrm{R}_{2}$ 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}} \Rightarrow \frac{\sigma_{1} 	imes 4 \pi \mathrm{R}_{1}^{2}}{4 \pi \varepsilon_{0} \mathrm{R}_{1}}=\frac{\sigma_{2} 	imes 4 \pi \mathrm{R}_{2}^{2}}{4 \pi \varepsilon_{0} \mathrm{R}_{2}}} $

${\Rightarrow \frac{\sigma_{1}}{\sigma_{2}}=\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}}$

Two charged spherical conductors of radii $R_1$ and $R_2$ are connected by a wire. The ratio of surface charge densities of the spheres $\sigma _1/\sigma _2$ will be

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