Question:
Consider two charged metallic spheres \(S_1\) and \(S_2\) of radii \(R_1\) and \(R_2\), respectively. The electric fields \(E_1\) (on \(S_1\)) and \(E_2\) (on \(S_2\)) on their surfaces are such that \(E_1/E_2 = R_1/R_2\). Then the ratio \(V_1\) (on \(S_1\)) / \(V_2\) (on \(S_2\)) of the electrostatic potentials on each sphere is :
Consider two charged metallic spheres \(S_1\) and \(S_2\) of radii \(R_1\) and \(R_2\), respectively. The electric fields \(E_1\) (on \(S_1\)) and \(E_2\) (on \(S_2\)) on their surfaces are such that \(E_1/E_2 = R_1/R_2\). Then the ratio \(V_1\) (on \(S_1\)) / \(V_2\) (on \(S_2\)) of the electrostatic potentials on each sphere is :
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For a charged sphere, surface potential \(V = ER\) (since \(E = kQ/R^2\), \(V = kQ/R\)). Take the ratio.
Updated On: Nov 5, 2026
- \((R_1/R_2)^2\)
- \((R_2/R_1)\)
- \((R_1/R_2)^3\)
- \(R_1/R_2\)
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The Correct Option is A
Solution and Explanation
For a charged metallic sphere, surface field \(E = \frac{kQ}{R^2}\) and surface potential \(V = \frac{kQ}{R}\), so \(V = ER\).
\(\frac{V_1}{V_2} = \frac{E_1 R_1}{E_2 R_2} = \frac{E_1}{E_2} \cdot \frac{R_1}{R_2} = \frac{R_1}{R_2} \cdot \frac{R_1}{R_2} = \left(\frac{R_1}{R_2}\right)^2\).
Answer: (1) \((R_1/R_2)^2\)
\(\frac{V_1}{V_2} = \frac{E_1 R_1}{E_2 R_2} = \frac{E_1}{E_2} \cdot \frac{R_1}{R_2} = \frac{R_1}{R_2} \cdot \frac{R_1}{R_2} = \left(\frac{R_1}{R_2}\right)^2\).
Answer: (1) \((R_1/R_2)^2\)
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