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A charged spherical conductor has a surface charge density sigma, and the electric field at its surface is E. If the radius of the sphere is doubled while keeping the surface charge density unchanged, what is the electric field at the surface of the new sphere?

1. E/4
2. E/2
3. E
4. 2E

Correct answer: E

Solution

The electric field just outside the surface of a charged conductor is E = sigma / epsilon₀, depending only on the surface charge density. Since sigma is unchanged, the field stays exactly the same E regardless of the radius. (Although total charge Q = sigma*4*pi*R² increases, the field expression in terms of sigma is radius-independent.)

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