Electrostatics, Energy of a uniformly charged sphere

NeoDevin
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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}, &amp;\mbox{ if } r &gt; R\\ \frac{kq}{R^3} r, &amp;\mbox{ if } r &lt; R \end{array}\right<br /> \\ \Rightarrow V(r) = \left\{\begin{array}{cc}\frac{kq}{r}, &amp; r &gt; R\\ \frac{3kq}{2R} - \frac{kq}{R^3} r^2, &amp; r &lt; 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 \\<br /> = \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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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
 
NeoDevin said:
Thanks, there's no sign errors, but I forgot a factor of 1/2 in the second term of V for r<R

I guess you're right. The sign error was mine!
 
Yeah, I worked it out with the correct expression for V, and it gives me the same answer now, thanks.
 
NeoDevin, you wouldn't happen to be in professor schick's class would you?
 
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