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Find the total electrostatic energy stored in the configuration

  1. Oct 10, 2014 #1
    1. The problem statement, all variables and given/known data
    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.

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

    3. The attempt at a solution
    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?
     
  2. jcsd
  3. Oct 10, 2014 #2

    vela

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    Staff Emeritus
    Science Advisor
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    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.
     
  4. Oct 10, 2014 #3
    Yeah I forgot about the parentheses. Have I done everything else right?
     
  5. Oct 11, 2014 #4
    can i get some help?
     
  6. Oct 12, 2014 #5

    gneill

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    Staff: Mentor

    Yes, it looks like you're doing okay setting up the integrals.
     
  7. Oct 12, 2014 #6
    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##.
     
  8. Oct 12, 2014 #7

    gneill

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    Staff: Mentor

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