Is this the correct set up for the electric field?

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
2 replies · 1K views
ghostfolk
Messages
58
Reaction score
1

Homework Statement


A spherical conductor of radius ##a## carries a charge q and also there is a jelly of constant charge ##rho## per unit volume extending from radius ##a## out to radius ##b##.
I'm looking to see if I got the correct set up for the electric field of this spherical conductor for all space.

Homework Equations


##\oint \vec{E} \cdot d\vec{a}=\frac{Q_{enc}}{\epsilon_0}##

The Attempt at a Solution


##\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}##
 
Physics news on Phys.org
Looks good except for the incorrect notation. You should write
$$\vec{E}=\begin{cases}
0 & r<a \\
\left[\rho \frac{(r^3-a^3)}{3r^2 \epsilon_0}+ \frac{q}{4\pi r^2\epsilon_0}\right] \hat{r} & a<r<b \\
\left[\rho \frac{(b^3-a^3)}{3r^2 \epsilon_0} +\frac{q}{4\pi r^2\epsilon_0}\right] \hat{r} & b\le r
\end{cases}.$$ 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.
 
vela said:
Looks good except for the incorrect notation. You should write
$$\vec{E}=\begin{cases}
0 & r<a \\
\left[\rho \frac{(r^3-a^3)}{3r^2 \epsilon_0}+ \frac{q}{4\pi r^2\epsilon_0}\right] \hat{r} & a<r<b \\
\left[\rho \frac{(b^3-a^3)}{3r^2 \epsilon_0} +\frac{q}{4\pi r^2\epsilon_0}\right] \hat{r} & b\le r
\end{cases}.$$ 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 should've added parentheses.Thanks.