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Total charge from charge density (spherical coordinates)

  1. Sep 17, 2013 #1
    1. The problem statement, all variables and given/known data
    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)?

    3. 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 im a little confused on how to take the integral with respect to dV in spherical coordinates. Im pretty sure i have to add an [itex] r^2 [/itex] in the integrand but im not sure.
  2. jcsd
  3. Sep 17, 2013 #2
    [itex] dV = r^2 \sin(\theta) dr d\theta d\phi[/itex]
  4. Sep 17, 2013 #3


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    Science Advisor

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

    Since you're using tex, just use \nabla for the del operator.
  5. Sep 17, 2013 #4
    lol ok, i didnt know that

    test [itex] \nabla [/itex]
  6. Sep 17, 2013 #5
    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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