Total charge from charge density (spherical coordinates)

bfusco
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Homework Statement


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

(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 [itex]\rho (\vec{r})[/itex] in (a). Does the answer agree with that in (b)?

The Attempt at a Solution


NOTE: IM USING [itex]\Delta[/itex] 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 [tex]\vec{ \Delta} \bullet \vec{E} = \frac{\rho}{\epsilon_0}[/tex]
rearranging it to solve for [itex]\rho[/itex] i get [tex]\rho = \epsilon( \vec{\Delta} \bullet \vec{E})[/tex] where [tex]\epsilon( \vec{\Delta} \bullet \vec{E}) = \frac{\epsilon_0}{r^2} \frac{d(r^2 k r^2)}{dr}[/tex] which reduces to [tex]\frac{r}{\pi}[/tex]

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

(c) knowing [itex]\rho = \frac{dq}{dV}[/itex] and reorganizing to solve for dq i get [itex]\int dq = \int \rho(\vec{r}) dV[/itex] 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 [itex]r^2[/itex] in the integrand but I am not sure.
 
on Phys.org
[itex]dV = r^2 \sin(\theta) dr d\theta d\phi[/itex]
 
bfusco said:
1
NOTE: IM USING [itex]\Delta[/itex] as the del operator


Don't! To most people,
[tex]\Delta = \nabla^2[/tex]

Since you're using tex, just use \nabla for the del operator.
 
phyzguy said:
Don't! To most people,
[tex]\Delta = \nabla^2[/tex]

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

lol ok, i didnt know that

test [itex]\nabla[/itex]
 
Bryson said:
[itex]dV = r^2 \sin(\theta) dr d\theta d\phi[/itex]

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

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