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We have a homogeneous charged ball with radius R which contains a ball-shaped hollow (with radius := r and distance from center of the bigger ball (M) to the center of the hollow (N) := b).

What’s now the field force inside the hollow, depending on b?


I first thought about calculating E(Ball R, without a hollow) - E(of the hollow r, displaced ) to use the superpostion principle, but then i would have no idea how to calculate the latter of those… would be happy about any hints!

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"... the field force.." Are you looking for the electric field, or a force the electric field exerts on some other object? Not to reprimand you but this is a common error students make - make sure you distinguish electric field and electric force. These are related but distinct concepts. – DJBunk Feb 22 '13 at 21:33
thank you for the correction! to precise: here i’m looking for the electric field… :) – user20486 Feb 22 '13 at 21:51
Couldn't you use the superposition principle, treating the hollow as a homogoneous charged ball with radius r with opposite charge density from the ball with radius R? – KDN Feb 23 '13 at 0:55
@user20486 - you might want to edit the question to reflect this, at the very least for people looking at the question in the future. Thanks. – DJBunk Feb 23 '13 at 20:51
up vote 0 down vote accepted

The answer is indeed the superposition principle. You have to calculate both the field of the ball with radius R and the field of a ball with radius r, placed away from the center at a distance b. Then you substract the field of the smaller ball from the larger one and you are done. This is equivalent to a superposition of the fields of two balls with charge of opposite sign.

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Thank you! This seems also for me the most reasonable way - this leads to the calculation and subtracting of those 2 electric fields (one standard electric field calculation for any point inside a homogeneous charged ball, and the same again just with smaller radius and displaced center with same charge density, where one just have to do a substitution) – user20486 Feb 24 '13 at 17:56

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