Normally it seems like this is solved by using Gauss's Law as follows:

$$ \int E * da= \frac {q_{enc}}{\epsilon_0} $$

And applying a geometric argument that says: because the charged sphere is uniform, we can take the ratio of the volume our Gaussian sphere as a function of some radius $r$ and divide it by the actual volume of the charged sphere, a ratio that gets multiplied to the total charge $Q$ for charge ratio. This returns:

$$ \frac {Qr^3}{R^3} $$

Which then gets plugged back into Gauss's Law for the final answer:

$$ E = \frac {Qr}{4\pi R^3\epsilon_0} $$

Where $r$ is the radial distance to a point inside the sphere and $R$ is the radius of the sphere. This all makes sense to me, but I was wondering if someone could explain how to solve this problem through an integration argument. I can't seem to find a resource that does, they fall back to this argument because it is easier, but I'd like to conceptually understand how to work it through integrating an expression for the charge to find $Q$ as a function of $r$ and then applying Gauss's Law.

  • $\begingroup$ You can do an integral, but the evaluation of the integral to prove that inside and outside shells behave differently is a bit tricky. $\endgroup$ Jan 31, 2017 at 7:21

1 Answer 1


Yes you can. As I recall, such comparison of two approaches was a must-have exercise in introductory EM course. In fact this result is one of the greatest idea of Newton, proved in Principia.

I suggest you read this: Shell theorem. The similar proof applies to electric force, And I shall leave it to you to convert gravity quantities to electric quantities.


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