JEE Mains · Physics · STD 12 - Electric potential and capacitance
- JEE Mains 2020
- Medium
- MCQ
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
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Answer & Solution
Correct Answer
(D) \(\left(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}\right)^{2}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{E}_{1}=\frac{\mathrm{KQ}_{1}}{\mathrm{R}_{1}^{2}}\)
\(\mathrm{E}_{2}=\frac{\mathrm{KQ}_{2}}{\mathrm{R}_{2}^{2}}\)
\(\frac{E_{1}}{E_{2}}=\frac{R_{1}}{R_{2}}\)
\(\cfrac{\frac{\mathrm{KQ}_{1}}{\mathrm{R}_{1}^{2}}}{\frac{\mathrm{KQ}_{2}}{\mathrm{R}_{2}^{2}}}=\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}\)
\(\frac{Q_{1}}{Q_{2}}=\frac{R_{1}^{3}}{R_{2}^{3}}\)
\(\frac{\mathrm{V}_{1}}{\mathrm{V}_{2}}=\frac{\mathrm{KQ}_{1} / \mathrm{R}_{1}}{\mathrm{KQ}_{2} / \mathrm{R}_{2}}=\frac{\mathrm{R}_{1}^{2}}{\mathrm{R}_{2}^{2}}\)
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