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## Homework Statement

Determine the total electrostatic potential energy of a nonconducting sphere of radius [itex]r_0[/itex] carrying a total charge [itex]Q[/itex] distributed uniformly thorughout its volume.

## Homework Equations

U = qV

## The Attempt at a Solution

[tex]\rho = \frac{Q}{4/3\pi r_0^3} = \frac{dq}{4\pi r^2dr}[/tex]

Electric field inside a nonconductor sphere

[tex]V = \frac{kQ}{2r_0}(3-\frac{r^2}{r_0^2})[/tex]

[tex]dU = Vdq[/tex]

[tex]= \frac{kQ}{2r_0} (3 - \frac{r^2}{r_0^2}) \rho4\pi r^2 dr[/tex]

[tex]= \frac{2kQ\pi\rho}{r_0}(3r^2 -\frac{r^4}{r_0^2} )dr[/tex]

After integrating it over [from zero to [itex]r_0[/itex]], I end up with the following result

[tex] \frac{3Q^2}{10\pi r_0 \epsilon_0}[/tex]

But the correct answer according to the textbook is this.

[tex]\frac{3Q^2}{20\pi r_0 \epsilon_0}[/tex]

It's almost the same result but missing a factor of two in the denominator. Is it because of the potential equation? Solutions manual uses potential at the surface, but in my answer I use potential inside a nonconductor. That may be the reason of 1/2. (kq/r_0 vs kq/

**2**r_0* (3 - r^2/r_0^2))