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## Homework Statement

If I have a solid sphere of radius R and charge density [itex]+\rho \, C/m^3[/itex] and I then remove a smaller sphere of radius [itex]b[/itex] and is a distance [itex]a[/itex] from the center of the larger sphere, what is the electric field inside the cavity?

I get an answer which I think is right. I've look at the math over and over and can't quite figure out what the teacher did.

## Homework Equations

Gauss' Law:

[tex] \oint \vec{E} \cdot d\vec{S} = \frac{q_{encl}}{\epsilon_0} [/tex]

## The Attempt at a Solution

I know this is a classic problem using the superposition principle.

First I apply GL to the large sphere at a distance [itex]r<R[/itex] from the center and get an electric field

[tex] \vec{E} = \frac{ \rho }{3 \epsilon_0} \vec{r} [/tex]

Then I do the same except with a charge density of [itex]-\rho \, C/m^3[/itex] so

[tex] \vec{E'} = - \frac{\rho }{3 \epsilon_0} \vec{r'} = -\frac{\rho}{3 \epsilon_0} (\vec{r} - \vec{a}) [/tex]

Now I just sum the two fields

[tex] \begin{align}

\vec{E} + \vec{E'} & = \frac{ \rho }{3 \epsilon_0} \vec{r} -\frac{\rho}{3 \epsilon_0} (\vec{r} - \vec{a}) \\ &= \frac{ \rho }{3 \epsilon_0} \vec{r} -\frac{\rho}{3 \epsilon_0} \vec{r} +\frac{\rho}{3 \epsilon_0} \vec{a} \\ &=\frac{\rho}{3 \epsilon_0} \vec{a} \end{align} [/tex]

This is my solution which is supposedly wrong.