Consider two charged metallic spheres S_{1} and mathrm{S}_{2} of radii mathrm{R}_{1} and mathrm{R}

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2. Electric Potential and Capacitance

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Consider two charged metallic spheres $S_{1}$ and $\mathrm{S}_{2}$ of radii $\mathrm{R}_{1}$ and $\mathrm{R}_{2},$ respectively. The electric $\left.\text { fields }\left.\mathrm{E}_{1} \text { (on } \mathrm{S}_{1}\right) \text { and } \mathrm{E}_{2} \text { (on } \mathrm{S}_{2}\right)$ on their surfaces are such that $\mathrm{E}_{1} / \mathrm{E}_{2}=\mathrm{R}_{1} / \mathrm{R}_{2} .$ Then the ratio $\left.\mathrm{V}_{1}\left(\mathrm{on}\; \mathrm{S}_{1}\right) / \mathrm{V}_{2} \text { (on } \mathrm{S}_{2}\right)$ of the electrostatic potentials on each sphere is

$\left(\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}\right)$

$\left(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}\right)^{3}$

$\left(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}\right)$

$\left(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}\right)^{2}$

(JEE MAIN-2020)

Solution

$\mathrm{E}_{1}=\frac{\mathrm{KQ}_{1}}{\mathrm{R}_{1}^{2}}$

$\mathrm{E}_{2}=\frac{\mathrm{KQ}_{2}}{\mathrm{R}_{2}^{2}}$

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

$\cfrac{\frac{\mathrm{KQ}_{1}}{\mathrm{R}_{1}^{2}}}{\frac{\mathrm{KQ}_{2}}{\mathrm{R}_{2}^{2}}}=\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}$

$\frac{Q_{1}}{Q_{2}}=\frac{R_{1}^{3}}{R_{2}^{3}}$

$\frac{\mathrm{V}_{1}}{\mathrm{V}_{2}}=\frac{\mathrm{KQ}_{1} / \mathrm{R}_{1}}{\mathrm{KQ}_{2} / \mathrm{R}_{2}}=\frac{\mathrm{R}_{1}^{2}}{\mathrm{R}_{2}^{2}}$

Standard 12

Physics

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Solution

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