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Question

Two charged conducting spheres of radii a and b are connected to each other by a wire. The ratio of electric fields at the surfaces of two spheres is:

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Solution

The correct option is **B** ba

Let a be the radius of a sphere A, Qa be the charge on the sphere, and Ca be the capacitance of the sphere.

Let b be the radius of a sphere B, Qb be the charge on the sphere, and Cb be the capacitance of the sphere. Since the two spheres are connected with a wire, their potential (V) will become equal.

Let Ea be the electric field of sphere A and Eb be the electric field of sphere B. Therefore, their ratio,

⟹EaEb=Qa4πϵ0a2×4πϵ0b2Qb

⟹EaEb=QaQb×b2a2

However, QaQb=Ca VCb V

And, CaCb=ab

∴QaQb=ab

⟹EaEb=abb2a2=ba

Therefore, the ratio of electric fields at the surface is ba

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