Total charge from charge density (spherical coordinates)

[bfusco]

Homework Statement

In some region of space, the electric field is \vec{E} =k r^2 \hat{r}, in spherical coordinates, where k is a constant.
(a) Use Gauss' law (differential form) to find the charge density \rho (\vec{r}).

(b) Use Gauss' law (integral form) to find the total charge containted in a sphere of radius R, centered at r=0

(c) Find the total charge contained in a sphere of radius R, centered at r=0 by direct integration of your result for \rho (\vec{r}) in (a). Does the answer agree with that in (b)?

The Attempt at a Solution

NOTE: IM USING \Delta as the del operator

Im pretty sure my problems only arise on part (c), but if any error is noticed in the other parts please tell me. Thank you

(a)using \vec{ \Delta} \bullet \vec{E} = \frac{\rho}{\epsilon_0}
rearranging it to solve for \rho i get \rho = \epsilon( \vec{\Delta} \bullet \vec{E}) where \epsilon( \vec{\Delta} \bullet \vec{E}) = \frac{\epsilon_0}{r^2} \frac{d(r^2 k r^2)}{dr} which reduces to \frac{r}{\pi}

(b) using \oint \Delta \bullet d\vec{a} = \frac{Q_e}{\epsilon_0} and rearranging to solve for Q_e i get Q_e=\epsilon_0 \oint \vec{E} \bullet d\vec{a}, which since the sphere is symmetric about the origin i can do \epsilon_0 \oint |\vec{E}| d\vec{a}, which equals \epsilon_0 k r^2 4\pi R^2 which reduces to R^4 (note: i replaced r^2 with R^2)

(c) knowing \rho = \frac{dq}{dV} and reorganizing to solve for dq i get \int dq = \int \rho(\vec{r}) dV at this point I am a little confused on how to take the integral with respect to dV in spherical coordinates. I am pretty sure i have to add an r^2 in the integrand but I am not sure.

 

[Bryson]

dV = r^2 \sin(\theta) dr d\theta d\phi

 

[phyzguy]

1
NOTE: IM USING \Delta as the del operator

Don't! To most people,
\Delta = \nabla^2

Since you're using tex, just use \nabla for the del operator.

 

[bfusco]

Don't! To most people,
\Delta = \nabla^2

Since you're using tex, just use \nabla for the del operator.

lol ok, i didnt know that

test \nabla

 

[bfusco]

dV = r^2 \sin(\theta) dr d\theta d\phi

alright so since non of these variables have any unit vector other than \hat{r} the r^2 sin\theta =0 ?? at which point the integral will be \oint \frac{r}{\pi} dr
which is \frac{r^2}{2\pi} which doesn't match part (b), and therefore i am incorrect somewhere.