Two hollow conducting spheres of radii R_1 and R_2 ( R_1 R_2 ) have equal charges. The potential would be: - Physics MCQ NEET 2022 2026 | ExamTest.live

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

Solution & Step-by-step Explanation

The electrostatic potential V on the surface of a conducting sphere of radius R with charge Q is: V = \frac{k Q}{R} Given that the charges Q are equal on both spheres: V \propto \frac{1}{R} Since R_{1} ≫ R_{2}, it implies \frac{1}{R _{2}} > \frac{1}{R _{1}}, so V_{2} > V_{1} .Potential will be more on the smaller sphere.

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

More on bigger sphere

More on smaller sphere

Equal on both the sphere

Dependent on the material properly of the sphere

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