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## Homework Statement

I have some difficulty understanding a part of the following solved problem:

Use Gauss's law to find the electric field inside a uniformly charged sphere (charge density ##\rho##).

Answer:

##\oint E.da = E.4 \pi r^2 = \frac{1}{\epsilon_0} Q_{enc} = \frac{1}{\epsilon_0} \frac{4}{3} \pi r^3 \rho##

##\therefore \ E= \frac{1}{3 \epsilon_0} \rho r \hat{r}##

I don't understand in the first step why the both the radius on the LHS and RHS are the same. I think the radius on the left should be that of the Gaussian surface and the one on the right should be that of the sphere itself...

## Homework Equations

The integral form of Gauss’s law:

##\oint_S \ \vec{E} . \hat{n} da = \frac{q_{enc}}{\epsilon_0}##

## The Attempt at a Solution

Here is a diagram of the Gaussian surface of radius R around the sphere of radius r:

http://img32.imageshack.us/img32/848/gaussian.jpg [Broken]

We have ##\rho=Q/Volume## for the entire sphere, so to get the Q needed for Gauss’s law we have to multiply it by the sphere's volume. So here is what I did:

##|E| \ \oint da= \frac{\rho.V}{\epsilon_0}##

##|E| 4 \pi R^2 = \frac{1}{\epsilon_0} \rho \frac{4}{3} \pi r^3##

##\therefore \ E= \frac{\rho r^3}{3 \epsilon_0 R^2}##

I chose the Gaussian surface to be larger than the sphere itself, so that's why the two radiuses have to be different and so they don't cancel. Am I wrong or is this a mistake in the book?

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