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A solid sphere of radius R carries a uniform volume charge. At what distance measured from its surface (outward) does the electrostatic potential drop to half of the potential at the centre?

1. R/3
2. R
3. R/2
4. 2R

Correct answer: R/3

Solution

For a uniformly charged solid sphere, V_centre = (3/2)(kQ/R). Half of this is (3/4)(kQ/R). Outside the sphere V = kQ/r; setting kQ/r = (3/4)(kQ/R) gives r = (4/3)R. The distance measured from the surface is therefore r - R = R/3.

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