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From the method of images we know that the potential everywhere above a grounded plate with a point charge above it is equal to

$$ V(x,y,z) = (\frac{q}{4 \pi \epsilon_0})[ \frac{1}{\sqrt{x^2 + y^2 + (z-d)^2}} - \frac{1}{\sqrt{x^2 + y^2 + (z+d)^2}}] $$

$$ = (\frac{q}{4 \pi \epsilon_0})[ \frac{1}{\sqrt{r^2 + d^2 - 2rd\cos(\theta)}} - \frac{1}{\sqrt{r^2 + d^2 + 2rd\cos(\theta)}}] $$

$$ = (\frac{q}{4 \pi \epsilon_0})[ \frac{1}{\sqrt{r^2 + d^2 - 2rd\cos(\theta)}} - \frac{1}{\sqrt{r^2 + d^2 - 2rd\cos(\theta')}}] $$

Where $\theta'$ is the angle between $r$ and the point $z = -d$.

To solve this problem with Laplaces equation and Legendre Polynomials we must express the problem in terms of a function $f$, with $\nabla^2f = 0$, and $V = V_0 + f$.

In this case, we know that $V$ must satisfy the condition that $V(r->\infty) = 0$, and that $V(\theta = \pi/2) = 0$. We know that, due to the free charge,

$$ V_0 = \frac{\lambda}{\sqrt{r^2 + d^2 - 2rd\cos(\theta)}} $$

thus, the boudary conditions on $f$ are:

$$ f(r->\infty) = 0$$

$$ f(\theta = \pi/2) = \frac{-\lambda}{\sqrt{r^2 + d^2}} $$

Where $\lambda = (\frac{q}{4 \pi \epsilon_0})$.

These conditions are satisfied the multipole expansion of $\frac{\lambda}{\sqrt{r^2 + d^2 - 2rd\cos(\theta)}}$:

$$ f(r, \theta) = (-\lambda)\sum_{l =0}^{\infty}\frac{r_<^l}{r_>^{l+1}}P_l(\cos\theta)$$

Where $r_<$ is less than $r_>$ are either $r$ or $d$.

However, we know that the solution to this problem is actually

$$ f(r, \theta) = (-\lambda)\sum_{l =0}^{\infty}\frac{r_<^l}{r_>^{l+1}}P_l(-\cos\theta) = (-\lambda)\sum_{l =0}^{\infty}(-1)^l\frac{r_<^l}{r_>^{l+1}}P_l(\cos\theta)$$

How do we derive the real answer from the boundary conditions?

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closed as off-topic by Kyle Kanos, Yashas, ZeroTheHero, Jon Custer, peterh Apr 30 '17 at 18:29

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You have the correct setup with the image charge, you just aren't doing the expansion correctly. For starters, with a grounded plate the definition of "grounded" is that the potential of the plate is 0, so: $$\Phi\left(\theta=\frac{\pi}{2}\right) = 0.$$ If the plate were at some other constant potential (and it must be constant since it's a conducting plate), that would be given in the problem and can be added to the grounded plate solution.

Next, we can get the needed expansions from your previous question, and apply it to the potential from the method of image charges: $$\begin{align} \Phi(\mathbf{r}, \mathbf{r}') &= \frac{q}{4\pi\epsilon_0} \left[\frac{1}{|\mathbf{r} - \mathbf{r}'|} - \frac{1}{|\mathbf{r} - \Pi_z \mathbf{r}'|}\right]\ \\ &\qquad (\Pi_z\ \mathrm{is\ the\ parity\ operator\ that\ reflects\ across\ the\ }z=0\mathrm{\ plane}) \\ & = \frac{q}{4\pi\epsilon_0} \left[\frac{1}{\sqrt{r^2 + d^2 - 2dr\cos\theta}}- \frac{1}{\sqrt{r^2 + d^2 - 2dr\cos(\pi - \theta)}}\right] \\ & = \frac{q}{4\pi\epsilon_0} \left[ \sum_{\ell=0}^\infty \frac{r_<^\ell}{r_>^{\ell+1}}P_\ell(\cos \theta) - \sum_{\ell=0}^\infty \frac{r_<^\ell}{r_>^{\ell+1}}P_\ell(-\cos \theta)\right] \\ & = \frac{2q}{4\pi\epsilon_0} \left[ \sum_{n=0}^\infty \frac{r_<^{2n+1}}{r_>^{2n+2}}P_{2n+1}(\cos \theta)\right]. \end{align}$$ That is, the potential is a sum over the odd Legendre polynomials because they are the ones that possess the correct parity (negative) under reflection across the $z=0$ plane.

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  • $\begingroup$ This works in this case, because we already know the position of the charge required to set the potential to be 0 across the conducting plate. However, in the event that we have no knowledge about where this charge should be (i.e., in the case that the configuration is actually of an infinite plate with a bump pointing away from the charge), we would not be able to do this. In other words, i'm attempting to find that there should be a charge at that position without just guessing and checking. $\endgroup$ – Ulad Kasach Apr 30 '17 at 0:43
  • $\begingroup$ Are you talking about a deformation of the plate [ "(i.e., in the case that the configuration is actually of an infinite plate with a bump pointing away from the charge)"], or a continuous charge distribution above the plate? $\endgroup$ – Sean E. Lake Apr 30 '17 at 0:45
  • $\begingroup$ Deformation of the plate. But that is different problem than the one I am asking about here. It is just that the problem of the deformed plate is only solvable in the method of finding an $f$ where $\nabla^2f = 0$ and $V = V_{due-to-charges} + f$ at all boundary conditions $\endgroup$ – Ulad Kasach Apr 30 '17 at 0:48
  • $\begingroup$ So, the potential due to charges is found using the method of images with $\Phi=0$ on the boundary, as we're doing here. That's what describes the way you can add a charge without disturbing the value of $\Phi$ on the boundary. This potential is the Green's function discussed here (and in Jackson) that is appropriate for Dirichlet boundary conditions. $\endgroup$ – Sean E. Lake Apr 30 '17 at 0:54
  • $\begingroup$ For some systems, like the problem of an infinite conductor with a circle cut out of it and a hemisphere of equal radius "welded" on the circle, the method of images is insufficient for deriving a problem. $\endgroup$ – Ulad Kasach Apr 30 '17 at 1:31

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