Determine the total electrostatic potential energy

[hitemup]

Homework Statement

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

Homework Equations

U = qV

The Attempt at a Solution

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

Electric field inside a nonconductor sphere
$$V = \frac{kQ}{2r_0}(3-\frac{r^2}{r_0^2})$$

$$dU = Vdq$$

$$= \frac{kQ}{2r_0} (3 - \frac{r^2}{r_0^2}) \rho4\pi r^2 dr$$
$$= \frac{2kQ\pi\rho}{r_0}(3r^2 -\frac{r^4}{r_0^2} )dr$$

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

$$\frac{3Q^2}{10\pi r_0 \epsilon_0}$$

But the correct answer according to the textbook is this.

$$\frac{3Q^2}{20\pi r_0 \epsilon_0}$$

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

 

[TSny]

This is a common mistake of "double counting". Suppose you had a system of 3 point charges. It is tempting to calculate the energy as $U = \sum_{i=1}^3q_iV_i$ where $V_i$ is the potential at the location of the i^th charge due to the other two charges. But if you write out the terms explicitly you will see that you are counting the interaction of each pair of charges twice.

 

[hitemup]

This is a common mistake of "double counting". Suppose you had a system of 3 point charges. It is tempting to calculate the energy as $U = \sum_{i=1}^3q_iV_i$ where $V_i$ is the potential at the location of the i^th charge due to the other two charges. But if you write out the terms explicitly you will see that you are counting the interaction of each pair of charges twice.

I can see that we've double counted the pairs in your example. But where exactly am I doing that in my solution?

 

[TSny]

In the integral $\int V dq$, $V$ represents the potential of the entire system at the point where $dq$ is located. So, consider two particular elements of charge $dq_1$ and $dq_2$. The integral will include contributions $V_1 dq_1$ and $V_2 dq_2$, where $V_1$ is the potential of the entire system at the location of $dq_1$ and $V_2$ is the potential of the entire system at $dq_2$.

Note that $V_1$ will contain a contribution from $dq_2$ of the form $\frac{k\; dq_2}{r_{12}}$ where $r_{12}$ is the distance between $dq_1$ and $dq_2$. So, the expression $V_1 dq_1$ contains a contribution of the form $\frac{k\; dq_2\; dq_1}{r_{12}}$, which is the potential energy of interaction between $dq_1$ and $dq_2$. But, by the same reasoning, you can see that $V_2 dq_2$ contains the same contribution again. So, the interaction between $dq_1$ and $dq_2$ is being counted twice.