# Electrostatics, Energy of a uniformly charged sphere

From Griffiths, Third edition Intro Electrodynamics

I've been working on this problem, and I get 2 different answers. I get the same answer for parts b and c, but not for a. I know I'm probably just doing something silly along the way, but I can't find my mistake...

## Homework Statement

Find the energy stored in a uniformly charged solid sphere of radius R and charge q. Do it three different ways:
(a) Use Eq. 2.43
(b) Use Eq. 2.45
(c) Use Eq. 2.44

## Homework Equations

Eq. 2.43 $$W=\frac{1}{2}\int \rho V d\tau$$
Eq. 2.45 $$W=\frac{\epsilon_{0}}{2}\int_{all space} E^2 d\tau$$
Eq. 2.44 $$W=\frac{\epsilon_{0}}{2}\left(\int_{V} E^2 d\tau + \oint_{S} VE\cdot da\right)$$

## The Attempt at a Solution

For part (a) we have:
$$E(r)=\left\{\begin{array}{cc}\frac{kq}{r^2}, &\mbox{ if } r > R\\ \frac{kq}{R^3} r, &\mbox{ if } r < R \end{array}\right \\ \Rightarrow V(r) = \left\{\begin{array}{cc}\frac{kq}{r}, & r > R\\ \frac{3kq}{2R} - \frac{kq}{R^3} r^2, & r < R \end{array}\right$$

Since it's uniformly charged, we know that:
$$\rho = \frac{3q}{4\pi R^3} \\$$

Then we can evaluate Eq. 2.43 to get:
$$W=2\pi\rho\int_{0}^R V(r) r^2 dr \\ = \frac{9}{20} \frac{kq^2}{R}$$

For (b) and (c) however I get:

$$W=\frac{3}{5} \frac{kq^2}{R}$$

So they are off by a factor of 3/4.

If you need to see more work to help me find my mistake, let me know and I will post more details.

Last edited:

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Dick
Homework Helper
E is the NEGATIVE gradient of the potential. You have a sign error. Also as written V is not even continuous across r=R, but I assume that is just a typo.

Thanks, there's no sign errors, but I forgot a factor of 1/2 in the second term of V for r<R

Dick