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I am thinking about a sphere of homogeneous charge distribution (see Electric field due to a solid sphere of charge, for example).

Setup: Charge imagined to be to the right of the light blue circle at a distance a from the center.

Here we use a Gaussian surface (blue area) to find the electric field $$ \epsilon_0 4\pi a^2 E = \frac{3Q_{sphere}}{4\pi R^3}\frac{4\pi a^3}{3} $$

I am told that if I were to put a test charge at $a=\frac{R}{2}$, it would feel a force of magnitude $F = qE$, $E$ calculated from above. This is something that intuitively doesn't make sense to me: If we think of the sphere as one that contains a lot of small charges $dq$, wouldn't all the $dq$'s to the right of the charge have an influence that would cancel some of the force exerted from the blue area? Sure, there is a green area at the other side of the blue area but the distance would mean that the electric field it generates would fall of and the situation wouldn't be symmetric.

Why doesn't the charge outside of the Gaussian surface contribute to the force?

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V.Vocor
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    See Newton's shell theorem. Essentially a duplicate of http://physics.stackexchange.com/q/18446/2451 , http://physics.stackexchange.com/q/2481/2451 and links therein, if electric charge and Coulomb's law are replaced by mass and Newton's law of universal gravitation, respectively. – Qmechanic Jun 06 '15 at 08:02
  • @Qmechanic why not just write a short answer that says this, before someone else does. – ProfRob Jun 06 '15 at 08:11

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