Can someone intuitively explain image charges to me?

Question

Ok, so here's what I understand from reading Purcell.

There's the uniqueness theorem. This theorem says that given some assortment of conductors with fixed potentials at their surfaces (boundary conditions), and $\nabla^2 \phi = 0$ everywhere in space (there's no point charges anywhere), then the solution for $\phi$ must be unique.

There's the simple example of a point charge $Q$ a distance $h$ over an infinite conducting plane of fixed potential $0$. This point charge can be approximated to be a conductor by approximating the charge being emitted from a very small sphere. Thus we have a system of conductors with bounds of $\phi_Q$ and $\phi_P=0$.

We can form a system with the "same boundaries" by placing a point charge $-Q$ a distance $-h$ from where the plane would be and leaving the point charge $Q$ a distance $h$ above where the plane would be. Now by virtue of $Q$ and $-Q$'s interactions, we have that between them is an equipotential line where the plane would be with potential $0$, mimicking $\phi_P = 0$, and the boundary condition of $\phi_Q$ remains the same.

Thus the boundary conditions are the same, and so the uniqueness theorem applies, and so that solution that applies to the $Q$ and $-Q$ system is the same solution as the one for the $Q$ - $Plane$ system. Thus we have the electric field everywhere above the plane.

But this really doesn't seem right. For one thing, it doesn't $seem$ like the boundary conditions are really satisfied. In the $Q$ and $-Q$ system, there are three boundary conditions right? The imaginary plane where the potential is $0$, the negative charge's boundary, and the positive charge's boundary. So this isn't really the same system. And if $-Q$ isn't counted as a conductor with a boundary condition, the Laplace's equation isn't satisfied anymore right? So we have to treat $-Q$ as a conductor with boundary conditions.

This example is a bit confusing. Can someone explain this and why image charges make sense to use at all? I also don't get why having a plane between both charges would not change the E-field. In my mind I imagine a certain E-field when the plane is there for a single charge $Q$, and then with two charges $Q$ and $-Q$ I imagine the E-field gets "narrower" because there is even more negative charge on the surface with the charge $Q$.

Answer 1

Wait a moment. You are getting confused with the boundary conditions and with the system of image charges.

Basically, you have to solve a boundary problem for the potential $\phi$. You are assuming this potential to be zero inside the conductor (not just on the surface), and non zero outside. You also have to supply another boundary condition: as you have a charge, the potential should diverge as $\frac{Q}{r}$ near the charge. You don't turn the charge in a conductor.

Now, image charges require you to remove the conductor, and place charges where the conductor was originally. You can do that because you already know the solution for $\phi$ inside the conductor, that is $\phi=0$: you can add and remove charges in this region freely, as you don't need the potential there.

When you remove the conductor and place the charge $-Q$ at $-h$, by calculating the potential on the plane that is orthogonal to the separation and exactly between the charges (where there was the conductor surface), you find out that the potential is zero on this surface. This is an hint of the fact that you can use the two charges solution in the charge+conductor system: basically, you solve in the $Q$ region for the potential, and then say that the potential inside the conductor is $0$ everywhere. At the end, you don't have the image charge anymore, just a constant potential. But the potential in the $Q$ region has exactly the right behaviour on the conductor's surface, so you can "glue" the solutions together.

EDIT: Let me be more precise, by precisely stating the problem in a mathematical form and understanding why the uniqueness theorem is well used here. To clarify, I'll use cgs units everywhere.

The problem becomes much more complicated when we have a conducting sphere instead of a charge, as in this case you would have that the charge on the sphere accumulates without spherical symmetry. Sure, in the limit that you can take of radius of the spheres going to zero things should get corrected, but it's an unnecessary complication and I think it is simply easier to learn to use charges as boundary conditions.

Let me explain the boundary conditions of the problem you have. You first have to specify the domain on which you are solving the problem. Let me set the plane at $z=0$ and the charge at $(0,0,h)$. In this case, you have to find the potential in the set $z>0$, removing the point $(x,y,z)=(0,0,h)$. We will indicate this set (positive half space without a point) as $\Sigma$. The boundary of this set is then given by the plane $z=0$, the point $(0,0,h)$ and the boundary at infinity, with $x^2+y^2+z^2\to\infty$ and $z>0$ (we will indicate this boundary as $\infty$). The boundary conditions are given by $\phi(z=0)=0$, $\phi(\infty)=0$ and, lastly, $\phi$ should go to $\frac{Q}{r}$ when you get near the sphere, with $r$ the distance from the charge. You have to take this boundary in the sense of a limit: the potential must diverge in $(0,0,h)$ as you have a point charge, and as you approach the charge the potential resembles more and more the potential of a single charge. Formally, you should have $\lim_{\vec{x}\to(0,0,h)}\phi(\vec{x})=\lim_{r\to0}\frac Qr$. If there is a different charge, you just substitute $Q$ with the appropriate charge.

Now, call $\vec{x}_+=(0,0,h)$ and $\vec{x}_-=(0,0,-h)$. The potential of the system of two point charges is given by $$ \Phi(\vec{x})=Q\left(\frac{1}{|\vec{x}-\vec{x}_+|}-\frac{1}{|\vec{x}-\vec{x}_-|}\right). $$ This potential solves Laplace's equation everywhere (but on the point charges) and goes to zero at infinity as a potential for a system of charges should do (that's self evident). For this potential, we have: $$ \Phi((x,y,0))=0,\quad\text{On the plane $z=0$ the potential is zero};\\ \lim_{|\vec x|\to\infty}\Phi(\vec x)=0,\quad\text{At infinity, the potential drops to zero};\\ \lim_{\vec x\to\vec x_+}\Phi(\vec x)=\lim_{r\to0}\frac Qr,\quad\text{Near the charge, the potential diverges as a point charge}. $$ Try to prove the first evaluation explicitly, it is a simple matter of substituting the argument. The second is trivial enough, while for the third you can argue that $\Phi$ has this boundary condition by writing $$ \lim_{\vec x\to\vec x_+}\Phi(\vec x)= \lim_{\vec x\to\vec x_+}\frac{Q}{|\vec x-\vec x_+|}\left(1+\frac{|\vec x-\vec x_+|}{|\vec x-\vec x_-|}\right). $$ Now, the second term in the parenthesis drops in the limit (as the denominator is different from zero, but the numerator is), so you get condition number 3. You also have the same divergence (with opposite sign) in $\vec x_-$, but we don't need that.

Now, we can say that if we equate the potential of the plane + charge system, $\phi(\vec x)$ to the potential of the two charges system $\Phi(\vec x)$ in $\Sigma$ (not everywhere), we have that $\phi(\vec x)$ has just the right limits when approaching to the boundaries of $\Sigma$. As $\Phi$ solves Laplace's equation in $\Sigma$ (and also for $z<0$, but we don't need that), $\phi$ also does. So $\phi$ has the right boundary conditions, and solves Laplace's equation: this proves that $\phi$ is THE solution for this system, due to uniqueness.

Lastly, let's build some physical intuition. At a conductor's surface, the field lines of the electric field become orthogonal to the surface of the conductor, and it abruptly goes to $0$ as soon as you cross the surface. This discontinuity in the electrical field (that is minus the derivative of the potential) determines the accumulated charge on the surface: the bigger the jump, the greater the charge accumulated at that point (sign depends on direction of the field). The two charge system has an electrical field in the $z=0$ plane that is most intense in the $x=0,y=0$ point, and then decreases with the distance from this point. As in $\Sigma$ the potential is given by $\Phi$ and, inside the conductor, the potential goes to zero and stays constant, the electric field is abruptly cut at the surface $z=0$. Most of the charge is concentrated in $(x,y)=0$, and then it decreases with the distance from this point (as the electrical field's discontinuity). The image charge method mimicks this charge distribution with an appropriate charge inside the conductor (that is then removed), and with a system of point charges the solution is trivial.