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Unformatted text preview: 49. (a) At A, the only ﬁeld contribution is from the +5.00Q particle in the hollow (this follows from
Gauss’ law – it is the only charge enclosed by a Gaussian spherical surface passing through point A,
concentric with the shell). Thus, using k for 1/4πε0 , we have E = k (5Q)/(0.5)2 = 20kQ directed
(b) Point B is in the conducting material, where the ﬁeld must be zero in any electrostatic situation.
(c) Point C is outside the sphere where the net charge at smaller values of radius is −3.00Q + 5.00Q =
2.00Q. Therefore, we have E = k (2Q)/(2)2 = 1 kQ directed radially outward.
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