Determine the total electrostatic potential energy

[hitemup]

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/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.