I was reading the method of images in Griffiths' electrodynamics book. He uses the classic image problem to explain the significance of the uniqueness theorem suppose a point charge $q$ is held a distance $d$ above an infinite grounded conducting plane the potential in the region above the plane can be calculated using method of image charge distribution it is a $z=0$ plane in 3D.

Here is what I understood:

Since the infinite conductive plane is grounded the potential is zero at the surface, which perfectly makes sense, which is one of the boundary conditions which is $V=0$ at $z=0$. The other boundary condition is $V$ approaches zero at infinity and this too makes sense. If we use Coulomb's law to find the potential at the surface, it would be $V = kq/d$; again this is not the solution since it does not satisfy the boundary condition that $V=0$ at $z-0$. To solve this problem, we introduce an image charge and solving for the potential for this new charge distribution we get an answer which satisfy both boundary conditions.

What I didn't understand:

The image charge is not real of course, but the obtained potential is the superposition of the potential due to the original charge $q$ and the image charge $-q$. To account for this situation, the theory goes like this: when we introduce the $q$ charge above the infinite grounded conducting plane, the negative charges ($-q$) are induced also exert the potential at the point P above the plane, but this completely contradicts the boundary condition which demands that $V=0$ at the surface. How can there be a situation where charges accumulated on the surface, which creates a definite electric field in normal direction and exerts a definite potential at point P above the plane (which we are concerned), have zero potential at the surface at the same time?

Shortly: How the infinite conducting plane still has a potential zero at the surface even after the accumulation of charges in surface due to the induction? does it mean the potential vanishes when the positive charge meets negative charge? doesn't the potential is negative infinity near the negative charge. How can it be zero?

Please explain it to me.


1 Answer 1


First an important general statement, although not so important for this question:

The method of image charges and currents does not always create the same potentials $\bf A$ and $V$ that you would get form integrating over the real charges (the induced surface charges and currents instead of the mirror images). It only guarantees that you get the same fields ($\bf E$ and $\bf B$).

A useful counter example is an AC current loop in the $xz$ plane, i.e. an oscillating magnetic dipole in the $y$-direction. Combined with an image it will only give some $\bf A$, with $V=0$ everywhere, while the real charges will give a different $\bf A$ and a nonzero $V$ (because there will be nonzero AC charge density on the surface!). The solutions for the potentials differ by just a gauge transformation, so the fields are the same.

Now to your questions:

  1. The negative charges ($-q$) that are induced do not "completely contradict the boundary condition" with their field, as you claim, because the boundary condition can only be applied to the total field of the charge $q$ together with the surface charges. You seem to (tacitly) assume that surface charge in an (infinitely) thin layer would give a very strong field, but that is not true, it just cancels the field of the charge $q$.
  2. How the plane still has a potential zero with the (real) induced charge can easily be computed, given that you can find the surface charge density is a simple expression: $$ \sigma(x,y) = \frac{-q\,h}{2\pi\,(x^2+y^2+h^2)^{3/2}} $$ so you can integrate it with the Green function $\frac1{4\pi\varepsilon_0\ |{\bf r}-{\bf r}'|}$ to find the potential. Just look in the center: $$\begin{align} V_\sigma(0,0) &= \int\limits_{-\infty}^\infty\!\! dx \int\limits_{-\infty}^\infty \!\! dy\ \frac{\sigma(x,y)}{4\pi\varepsilon_0\,\sqrt{x^2+y^2}} \\[6pt] &=\int\limits_{-\infty}^\infty\!\! dx \int\limits_{-\infty}^\infty \!\! dy\ \frac{-q\,h}{2\pi\,(x^2+y^2+h^2)^{3/2}\ 4\pi\varepsilon_0\,\sqrt{x^2+y^2}} \\[6pt] &=\int\limits_{0}^\infty\!\! 2\pi\,d\rho\ \frac{-q\,h}{2\pi\,(\rho^2+h^2)^{3/2}\ 4\pi\varepsilon_0\,\sqrt{\rho^2}} = \frac{-q}{4\pi\varepsilon_0 \,h}, \end{align}$$ so that is clearly a finite result and it is also exacly the opposite of the potential of the charge $q$. So no problems here!
  • $\begingroup$ so the plane with induced charges can create a potential at point P but itself has zero potential? I didnt understand that point. Although it is grounded it still can be charged I understand that but if the induced charge creates a same potential as that of the original q the potential must be zero everywhere right? If both charges are equal. I didn't understand that part $\endgroup$
    – Hello
    Commented May 19 at 12:56
  • $\begingroup$ But both charges are not equal. They are opposite! So if one creates a potential the other can create the opposite potential. (Not trivial, since the $+q$ charge is a point, and the countercharge is spread out in some distribution over a plane, but at least there's no reason to think it's impossible.) $\endgroup$ Commented May 19 at 13:50
  • $\begingroup$ Yet sir how it still has zero potential? So if I bring the test charge(-q) from infinity to the region near the surface of the plane where the distribution of induced charges are , No work has to be done? Even though it produces a definite potential? I'm not saying it is not possible sir I understand but even if we consider the net effect the accumulated charges still contributes. How it is possible sir? $\endgroup$
    – Hello
    Commented May 19 at 14:19
  • $\begingroup$ No, if you bring in the charges it means they are not yet there, so the potential is not yet zero. The first charges you "bring in" feel the full potential of the point charge $+q$. Later, after some negative charge is already there, the potential starts to diminish, and the very last bit of negative charge will feel almost no potential any more; if you bring it in the job is almost done. But you have to look at the whole job! $\endgroup$ Commented May 19 at 14:23
  • $\begingroup$ so if I understand correctly sir. What you are saying is the potential of positive charge on test charge is cancelled by the accumulated negative charge? If that is the case near the surface the negative test charge would experience large amount of negative potential and pushed back to make a stable configuration the work has to be done there. And even if we consider total work done it would still be non zero wouldn't? Correct me if I'm wrong sir. $\endgroup$
    – Hello
    Commented May 19 at 14:37

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