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Question

# Two conducting spheres of radii a and b respectively are charged and joined by a wire. The ratio of electric field due the spheres at their surfaces when isolated is

A
ab
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B
ba
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C
a2b2
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D
b2a2
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Solution

## The correct option is B baJoined by a wire means they are at the same potential. For same potential kQ1a1=kQ2a2⇒Q1Q2=ab Further, the electric field at the surface of the sphere having radius R and charge Q is kQR2 ∴E1E2=kQ1a2kQ2b2=Q1Q2×b2a2=ba

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