Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Image Charges Purcell Discussion

  1. Jul 16, 2016 #1
    Hi everyone, I have a question (or three) regarding Purcell's discussion of image charges, in pg 137-139 of the third edition of his EM book. He goes over the classic image charge example: a plane held at zero potential, and a charge [itex] Q [/itex] a distance [itex] h [/itex] above it.
    I'm fine with everything he says, up until the following: "returning to the actual setup with the conducting plane, we know that in terms of the surface charge density [itex] \sigma [/itex], the electric field just above the plane is [itex] E_z = \sigma / \epsilon_0 [/itex]." My problem with this is that he then proceeds to plug in the expression for [itex] E_z [/itex] that he had determined from the image charges situation. But this [itex] E_z [/itex] corresponds to the electric field due to the whole system, charge [itex] Q [/itex] included. So why would [itex] E_z [/itex] correspond to just the contribution from the plane, [itex] \sigma / \epsilon_0 [/itex]?

    That's my first question. My second question stems from the following; having found the charge distribution on the plane, he integrates through the whole plane to find that the total charge on it is [itex] -Q [/itex], and explains that even if the plane initially had zero charge, this makes sense. The justification he gives is that a compensating positive charge [itex] Q [/itex] is spread over the whole plane, but that its contribution to the electric field [itex] E_z [/itex] is negligible (since the charge density will be zero in the infinite plane). I have the following problem with this argument:
    It seems like the charge distribution [itex] \sigma [/itex] should somehow be balanced by the positive charge, so that it integrates to zero at infinity. Otherwise, it's not the actual (real, physical) charge distribution! (Since the plane had zero charge to begin with.) It's just the approximate distribution that we find at points near the origin of the plane (below the charge [itex] Q [/itex].)


    Then I have an extra bonus question :)
    Purcell argues that the field below the plane must be zero, because "we can consider the conducting plane to be the top of a very large conducting sphere, and we know that the field inside a conductor is zero." But I have a huge problem with this. If we consider the plane to be that (the top of a sphere), then we have different boundary conditions from before! So it becomes a little sketchy, in my opinion, to use this argument, and combine it with the image charges he used (which definitely don't satisfy the boundary conditions of zero potential along the whole "big sphere" that Purcell is imagining; it just had zero potential along the infinite plane). How is this justified? Can anybody give another argument why the field below the plane must be zero?


    As you can probably tell, I had a lot of trouble following this discussion. Any help would be appreciated!
     
  2. jcsd
  3. Jul 17, 2016 #2

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    The point is to solve Maxwell's equations with the appropriate boundary conditions. The image charge is just a heuristic argument to find that solution. It is not really present. Physically what happens putting the charge ##q## at ##(0,0,h)## is that the charges in the metal plane rearrange due to the force acting on them due to the charge ##q##. In the static situation they must arrange themselves such that the total force on them is 0.

    It's clear that the field is ##0## for ##z<0## since this for sure fulfills the boundary conditions. For ##z>0## we get the field by superposition of the Coulomb field of the external charge at ##(0,0,h)## and that of the influence charge on the plane. Now we get the correct field for ##z>0## by symmetry introducing the fictitious image charge, i.e., the electrostatic potential is (in Heaviside-Lorentz units)
    $$\phi(\vec{x})=\begin{cases}
    0 & \text{for} \quad z<0,\\
    \frac{q}{4 \pi} \left (\frac{1}{\sqrt{x^2+y^2+(z-h)^2}} - \frac{1}{\sqrt{x^2+y^2+(z+h)^2}} \right) & \text{for} \quad z>0.
    \end{cases}$$
    the influence-charge distribution on the plane is given by the normal component of the electric field, ##\vec{E}=-\vec{\nabla} \phi##, and Maxwell's equation, ##\vec{\nabla} \cdot \vec{E}=4 \pi \rho## tells you that it's the total electric field at this plane!). Indeed at the plane (at ##z=0^+##) the field is
    $$\vec{E}(x,y,0^+)=-\frac{h q}{2 \pi (x^2+y^2+h^2)^{3/2}} \vec{e}_z=\sigma \vec{e}_z.$$
    The total influence charge is given by Gauss's theorem without further calculation. Just take a large cube with one boundary on the ##xy## plane and then taking the limit of an infinitely large cube. All the boundaries except the one on the ##xy## plane are at infinity then, and the field vanishes there. So you really only integrate the surface-charge density over the entire ##xy## plane. For the volume integral we note that the field effectively integrated over is again given by the field of the external charge and the virtual image charge. Only the image charge is inside the infinitely large cube and, thus the integral must be ##-q##.

    You can get this of course also by integrating the surface-charge density, using polar coordinates
    $$Q_{\text{surface}}=\int_0^{\infty} \mathrm{d} \rho \int_0^{2 \pi} \mathrm{d} \varphi \rho \sigma(\rho)=-\int_0^{\infty} \mathrm{d} \rho \frac{h q \rho}{(\rho^2+h^2)^{3/2}} = \left. \frac{h q}{\sqrt{\rho^2+h^2}} \right|_{\rho=0}^{\infty} = -q.$$
     
  4. Jul 17, 2016 #3
    Thanks for your answer! To answer my first question, I see that you can just use a Gaussian surface around the plane and the charge distribution is easily found in terms of the electric field.
    Unfortunately I understood most of the other parts of the argument already (it's similar to what Purcell said)- my other main questions, 2 and 3, still stand. Do you have any input on those?
     
  5. Jul 17, 2016 #4
    Now that I think about it, I have an idea of what might be going on, though I'm not sure.
    My problem was this: if we have the plane at zero potential, and zero charge, and bring the positive charge near the plane, it shouldn't mean that somehow now the plane has acquired a negative charge, and usually, this doesn't happen. However, if we impose that the plane remain at zero potential throughout this whole process, maybe that implies that we have to keep it connected to an outside source (to keep it grounded); if this is the case, then maybe positive charge could leak out of the plane (or negative charge leak in, equivalently) as we bring the positive point charge closer to the plane. Maybe this is the reason the plane supposedly has charge [itex] -Q [/itex] on it after this whole process.
     
  6. Jul 18, 2016 #5

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    In my opinion I answered all your questions already. The given solution is unique since the boundary-value problem of the Poisson equation is unique.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Image Charges Purcell Discussion
  1. Image charge (Replies: 5)

  2. Image charges (Replies: 2)

  3. Image charge problem (Replies: 1)

Loading...