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Induced Charge on a Grounded Sphere

  1. Sep 21, 2016 #1
    1. The problem statement, all variables and given/known data
    A point charge [itex]q[/itex] is located a distance [itex]d[/itex] away from the centre of a grounded conducting sphere of radius [itex]R<d[/itex]. I need to find the charge density on the sphere and the total induced charge on the sphere.
    This is very similar to example 2 here; http://web.hep.uiuc.edu/home/serrede/P435/Lecture_Notes/P435_Lect_06.pdf

    2. Relevant equations
    Gauss's Law in differential from; [itex]\rho=\epsilon_0(\nabla \cdot E)[/itex].

    The electric field in terms of potential; [itex]E=-\nabla V[/itex]

    Potential for a point charge; [itex] V=\frac{1}{4\pi\epsilon_0}\frac{q}{|\vec{r}|} [/itex]

    3. The attempt at a solution
    First I found the potential and obtained the same result as in the provided link, that is; [tex]
    V(r,\theta)=\frac{q}{4\pi\epsilon_0}\left [ \frac{1}{\sqrt{d^2+r^2-2drcos(\theta)})}-\frac{R}{d\sqrt{(R^2/d^2)+r^2-2(R^2/d)rcos(\theta)}} \right ] [/tex].
    Using an image charge of magnitude [itex]-\frac{R}{a}q[/itex] located at [itex]\frac{R^2}{d}[/itex]

    Next I try to find the electric field via; [tex]E=-\nabla V=\left ( -\frac{\partial V}{\partial r},-\frac{1}{r}\frac{\partial V}{\partial \theta} \right )[/tex]
    This is where things start to get really messy (the derivatives are simple so i wont type them out, they're just really messy). Also, this is in spherical coordinates.
    Finally, I use this to find the charge density via Gauss's Law. Applied to the above, this yields;
    [tex]
    \rho=\epsilon_0\frac{1}{r^2}\frac{\partial}{\partial r}\left ( -r^2\frac{\partial V}{\partial r} \right )+\epsilon_0\frac{1}{rsin(\theta)}\frac{\partial}{\partial \theta}\left ( \frac{-sin(\theta)}{r}\frac{\partial V}{\partial \theta} \right )
    [/tex]
    And then evaluate this at [itex]r=R[/itex] to find the charge density on the sphere.

    This expression ends up being incredibly nasty. As I also have to find the total induced charge I would then need to integrate this expression over the sphere. By rights this should give an induced charge of [itex]-q[/itex]. However, I have no idea how to uintegrate what I obtain, and Mathematica gives a different answer (which happens to be in terms of elliptic integrals).

    So, my question has three parts. Firstly, is my reasoning/understanding in the above correct?
    Secondly, in the provided link they claim that the charge density is simply given by; [tex]\rho=-\epsilon_0\frac{\partial V}{\partial r}[/tex]
    Evaluated at the radius of the sphere. I have no idea how this is derived (I've seen it used a fair bit while searching around, but I dont understand how it is derived or where it is applicable).
    Finally, can anybody point me in the right direction?
    Thanks in advance for any help!
     
  2. jcsd
  3. Sep 21, 2016 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Hello. Welcome to PF!

    ##\rho## in Gauss' law represents the volume charge density. But a conductor in electrostatic equilibrium will only have charge on the surface. So, you should try to find the surface charge density ##\sigma##.
    Here, ##\rho## should be ##\sigma##. To understand the relation between ##\sigma## and ##\frac{\partial V}{\partial r}## you should recall some basic facts about conductors in electrostatic equilibrium. In particular, you should be able to answer the following:
    1. What is the direction of the electric field at the surface of the conductor?
    2. What is the relation between the surface charge density and the electric field at the surface?
     
  4. Sep 22, 2016 #3
    OHHHH, thank you! I knew I must have been overlooking something simple. Near the surface of a conductor the electric field has to be perpendicula to said surface (in equilibrium). Then you can note that the electric field inside the conductor is zero, apply Gauss's law in integral form and there's the result! Thanks for your help, I really appreciate it!

    EDIT:

    Sorry for asking so many questions, but now that I understand the first part of my propblem I can't seem to figure out the second half (finding the total charge induced).
    By applying [itex]\sigma=-\epsilon_0\frac{\partial V}{\partial r}=\epsilon_0E_r[/itex] to the potential above, I find that
    [tex]
    \sigma=-\epsilon_0\frac{\partial V}{\partial r}=\frac{q}{4\pi}\left [ \frac{R-dcos(\theta)}{(d^2+R^2-2dRcos(\theta))^{3/2}}-\frac{d-Rcos(\theta)}{d^2R(R^2/d^4-(2R/d)cos(\theta)+1)^{3/2}} \right ]
    [/tex]
    Which I'm pretty sure is right. When I plot this against [itex]\theta[/itex] it looks as expected, high charge density for small values (the near side of the sphere), and decreasing in magnitude with theta.
    To find the total charge I think I'll need to integrate this function over the surface of the sphere. As far as I can tell (I'm still pretty new to multivariate calculus), this will be
    [tex]\int_{0}^{2\pi}\int_{0}^{\pi}\frac{qR^2sin(\theta)}{4\pi}\left [ \frac{R-dcos(\theta)}{(d^2+R^2-2dRcos(\theta))^{3/2}}-\frac{d-Rcos(\theta)}{d^2R(R^2/d^4-(2R/d)cos(\theta)+1)^{3/2}} \right ]d\theta d\phi
    [/tex]
    This seems really hard to compute though, so Im not sure if Ive made another mistake, or if I've maybe come up with the wrong integral. This should come out to be the negative of our image charge, but Mathematica gives an expression involving a bunch of square roots and stuff, which is definitely not the expectd result.
    Sorry for all the questions, but thanks for all the help!
     
    Last edited: Sep 22, 2016
  5. Sep 22, 2016 #4

    TSny

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    Homework Helper
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    Look at the second term:## \frac{d-Rcos(\theta)}{d^2R(R^2/d^4-(2R/d)cos(\theta)+1)^{3/2}}##. The first term inside the (... )3/2 does not have the right dimensions.

    I looked back at your first post where you gave an expression for V:
    [tex]
    V(r,\theta)=\frac{q}{4\pi\epsilon_0}\left [ \frac{1}{\sqrt{d^2+r^2-2drcos(\theta)})}-\frac{R}{d\sqrt{(R^2/d^2)+r^2-2(R^2/d)rcos(\theta)}} \right ] [/tex].

    There is a similar dimensional analysis problem with the second term.

    After taking the derivative of V with respect to ##r## and evaluating at ##r = R##, you should be able to simplify quite a bit by getting the two (... )3/2 factors to be identical.
     
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