Find the total electrostatic energy stored in the configuration

[ghostfolk]

Homework Statement

A spherical conductor of radius $a$ carries a charge $q$ and also there is a jelly of constant charge density $\rho$ per unit volume extending from radius a out to radius $b$. Find the electrostatic energy stored in the configuration.

Homework Equations

$\oint \vec{E} \cdot d\vec{a}=\frac{Q_{enc}}{\epsilon_0}$
$U=\frac{\epsilon_0}{2}\int E^2 d^3r$

The Attempt at a Solution

[/B]We first find the electric field.
$\oint \vec{E} \cdot d\vec{a}=4\pi r^2$
$Q_{enc}=\int_a^r 4\pi r'^2 \rho dr'+q=\frac{4\pi}{3}(r^3-a^3)\rho+q$
So then,
$E=\begin{cases} 0, r<a& \\\ \rho \frac{(r^3-a^3)}{3r^2 \epsilon_0}+ \frac{q}{4\pi r^2\epsilon_0} \hat{r}, a<r<b\\ \rho \frac{(b^3-a^3)}{3r^2 \epsilon_0} +\frac{q}{4\pi r^2\epsilon_0} \hat{r}, b\le r \end{cases}$
Then
$E^2=\begin{cases} 0, r<a& \\\ \rho^2 \frac{(r^3-a^3)^2}{9r^4 \epsilon_0^2}+ 2q \rho \frac{(r^3-a^3)}{12 \pi r^4\epsilon_0^2}+\frac{q}{16 \pi^2 r^4 \epsilon_0^2}, a<r<b\\ \rho^2 \frac{(b^3-a^3)^2}{9r^4 \epsilon_0^2}+ 2q \rho \frac{(b^3-a^3)}{12 \pi r^4\epsilon_0^2}+\frac{q}{16 \pi^2 r^4 \epsilon_0^2}, b\le r \end{cases}$
Now,
$U=\frac{\epsilon_0}{2} \int_a^b (\rho^2 \frac{(r^3-a^3)^2}{9r^4 \epsilon_0^2}+ 2q \rho \frac{(r^3-a^3)}{12 \pi r^4\epsilon_0^2}+\frac{q}{16 \pi^2 r^4 \epsilon_0^2})4 \pi r^2dr+\frac{\epsilon_0}{2} \int_b^\infty (\rho^2 \frac{(b^3-a^3)^2}{9r^4 \epsilon_0^2}+ 2q \rho \frac{(b^3-a^3)}{12 \pi r^4\epsilon_0^2}+\frac{q}{16 \pi^2 r^4 \epsilon_0^2}) 4 \pi r^2dr$

So far, have I done everything correctly?

 

[vela]

No, your expression for the electric field is non-sensical. It should be
$$\vec{E}=\begin{cases}
0 & r<a \\
\left[\rho \frac{(r^3-a^3)}{3r^2 \epsilon_0}+ \frac{q}{4\pi r^2\epsilon_0}\right] \hat{r} & a<r<b \\
\left[\rho \frac{(b^3-a^3)}{3r^2 \epsilon_0} +\frac{q}{4\pi r^2\epsilon_0}\right] \hat{r} & b\le r
\end{cases}.$$ The way you wrote it, $\hat{r}$ only multiplies the last term, and you'd be adding a scalar to a vector, which doesn't make sense.

 

[ghostfolk]

No, your expression for the electric field is non-sensical. It should be
$$\vec{E}=\begin{cases}
0 & r<a \\
\left[\rho \frac{(r^3-a^3)}{3r^2 \epsilon_0}+ \frac{q}{4\pi r^2\epsilon_0}\right] \hat{r} & a<r<b \\
\left[\rho \frac{(b^3-a^3)}{3r^2 \epsilon_0} +\frac{q}{4\pi r^2\epsilon_0}\right] \hat{r} & b\le r
\end{cases}.$$ The way you wrote it, $\hat{r}$ only multiplies the last term, and you'd be adding a scalar to a vector, which doesn't make sense.

Yeah I forgot about the parentheses. Have I done everything else right?

 

[ghostfolk]

can i get some help?

 

[gneill]

So far, have I done everything correctly?

Yes, it looks like you're doing okay setting up the integrals.

 

[ghostfolk]

Yes, it looks like you're doing okay setting up the integrals.

Do you think there could've been an easier method? I know there's the other equation $U=\int \rho V d^3r$, but I felt that it would be just as tedious being that I would still need to find the electric field to get $V$.

 

[gneill]

Do you think there could've been an easier method? I know there's the other equation $U=\int \rho V d^3r$, but I felt that it would be just as tedious being that I would still need to find the electric field to get $V$.

I don't think you'll be able to avoid messy integrals either way.