- #1

erogard

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## Homework Statement

Hi everyone,

I'm supposed to find an expression for the electric potential as a function of r, the radial distance inside a solid and non conducting sphere of radius R. A total charge of q is uniformly distributed throughout its volume. The annoying part is that I'm supposed to do that using a relation between the potential and [tex] \rho[/tex] , the charge density.

## Homework Equations

The equation the problem wants me to use is:

[tex] V( \textbf{r} ) = \frac{1}{4 \pi \epsilon_{0}} \int \frac{\rho ( \textbf{r} )}{r} d \tau ' [/tex]

[tex]r[/tex] being the distance between source & field (observation) point.

I already have obtained a result using a different method (integrating the electric field) and got:

This is also what I found on different sources so I'm pretty sure the result is correct.

## The Attempt at a Solution

Yet I can't set up my integration is such a way so as to retrieve this result.

Using the law of cosines on the following figure:

I have: [tex] r^{2} = r'^{2} + z^{2} - 2 r' z cos ( \theta )[/tex]

Using spherical coordinates, I make the integral range from 0 to 2pi and 0 to pi for phi and theta.

For [tex]r'[/tex] I am more doubtful. Indeed, when we integrate with respect to the two angles first, we obtain an expression that involves the square root of the following difference: [tex](r' - z)^{2}[/tex].

Thus depending on where the observation point is on the sphere, the positive or negative root is to be taken. So I would guess I could split the integration into two parts, from R (the radius of the sphere) to r, and then from r to 0. Since [tex]\rho[/tex] is zero everywhere outside the sphere, there is no need to integrate from infinity, where the potential is assumed to go to zero.

Yet I do not get the expected result, and this using different integration limits. I am assuming I am doing something wrong initially, and if someone has any suggestion I would be very grateful for that.

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