Two hollow conducting spheres of radii R_{1} | vedclass

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(A) The electric potential $V$ at the surface of a hollow conducting sphere of radius $R$ carrying charge $Q$ is given by the formula: $V = \frac{1}{4

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(A) The electric potential $V$ at the surface of a hollow conducting sphere of radius $R$ carrying charge $Q$ is given by the formula: $V = \frac{1}{4 \pi \epsilon_{0}} \cdot \frac{Q}{R}$.Here,$\frac{1}{4 \pi \epsilon_{0}}$ is a constant.It is given that both spheres have the same charge $Q$.Therefore,the potential $V$ is inversely proportional to the radius $R$,i.e.,$V \propto \frac{1}{R}$.Since $R_{1} >> R_{2}$,the radius of the smaller sphere is $R_{2}$.Because $R_{2}  V_{1}$.Thus,the potential is higher on the smaller sphere.

Two hollow conducting spheres of radii $R_{1}$ and $R_{2}$ $(R_{1} >> R_{2})$ have equal charges. The potential would be:

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