Two charged spherical conductors of radius R_ | vedclass

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(B) When two conductors are connected by a wire,charge flows until their potentials become equal.Let $V_{1}$ and $V_{2}$ be the potentials of the sphe

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$\frac{R_{2}}{R_{1}}$

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(B) When two conductors are connected by a wire,charge flows until their potentials become equal.Let $V_{1}$ and $V_{2}$ be the potentials of the spheres. Since they are connected,$V_{1} = V_{2}$.The potential of a charged spherical conductor is given by $V = \frac{kQ}{R}$.Thus,$\frac{kQ_{1}}{R_{1}} = \frac{kQ_{2}}{R_{2}}$,which implies $\frac{Q_{1}}{Q_{2}} = \frac{R_{1}}{R_{2}}$.The surface charge density $\sigma$ is defined as $\sigma = \frac{Q}{A} = \frac{Q}{4\pi R^{2}}$.Therefore,the ratio of surface charge densities is:$\frac{\sigma_{1}}{\sigma_{2}} = \frac{Q_{1} / (4\pi R_{1}^{2})}{Q_{2} / (4\pi R_{2}^{2})} = \frac{Q_{1}}{Q_{2}} 	imes \frac{R_{2}^{2}}{R_{1}^{2}}$.Substituting $\frac{Q_{1}}{Q_{2}} = \frac{R_{1}}{R_{2}}$,we get:$\frac{\sigma_{1}}{\sigma_{2}} = \frac{R_{1}}{R_{2}} 	imes \frac{R_{2}^{2}}{R_{1}^{2}} = \frac{R_{2}}{R_{1}}$.

Two charged spherical conductors of radius $R_{1}$ and $R_{2}$ are connected by a wire. Then the ratio of surface charge densities of the spheres $(\sigma_{1} / \sigma_{2})$ is:

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