Consider two charged metallic spheres S_{1} a | vedclass

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(D) The electric field on the surface of a charged sphere is given by $E = \frac{KQ}{R^{2}}.$Given $E_{1} = \frac{KQ_{1}}{R_{1}^{2}}$ and $E_{2} = \fr

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$(R_{1} / R_{2})^{2}$

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(D) The electric field on the surface of a charged sphere is given by $E = \frac{KQ}{R^{2}}.$Given $E_{1} = \frac{KQ_{1}}{R_{1}^{2}}$ and $E_{2} = \frac{KQ_{2}}{R_{2}^{2}}.$According to the problem,$\frac{E_{1}}{E_{2}} = \frac{R_{1}}{R_{2}}.$Substituting the expressions,we get $\frac{KQ_{1} / R_{1}^{2}}{KQ_{2} / R_{2}^{2}} = \frac{R_{1}}{R_{2}}.$$\frac{Q_{1}}{Q_{2}} \cdot \frac{R_{2}^{2}}{R_{1}^{2}} = \frac{R_{1}}{R_{2}} \implies \frac{Q_{1}}{Q_{2}} = \frac{R_{1}^{3}}{R_{2}^{3}}.$The electrostatic potential on the surface of a sphere is $V = \frac{KQ}{R}.$Therefore,$\frac{V_{1}}{V_{2}} = \frac{KQ_{1} / R_{1}}{KQ_{2} / R_{2}} = \frac{Q_{1}}{Q_{2}} \cdot \frac{R_{2}}{R_{1}}.$Substituting $\frac{Q_{1}}{Q_{2}} = \frac{R_{1}^{3}}{R_{2}^{3}},$ we get $\frac{V_{1}}{V_{2}} = \frac{R_{1}^{3}}{R_{2}^{3}} \cdot \frac{R_{2}}{R_{1}} = \frac{R_{1}^{2}}{R_{2}^{2}} = (R_{1} / R_{2})^{2}.$

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