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A uniformly charged non-conducting solid sphere of radius R carries total charge Q. A narrow tunnel is drilled through the sphere along a diameter. A point charge q of mass m is placed at one end of the tunnel (on the surface of the sphere, point A) and released. What is the minimum initial speed v0 that must be given to the charge so that it reaches the other end B (diametrically opposite point on the surface)?
- v0 = sqrt(2KQq / (4mR))
- v0 = 0
- v0 = sqrt(6KQq / (4mR))
- v0 = sqrt(3KQq / (4mR))
Correct answer: v0 = 0
Solution
If q is of the same sign as Q, it is repelled and the centre (highest potential) is the hardest point. But if q is opposite in sign to Q, the centre is a potential energy minimum and B (surface) has the same potential as A, so no minimum speed is needed. For diametrically opposite identical surface points, V(A) = V(B), and hence zero minimum speed is required.
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