Consider the typical problem,

You have a conducting sphere of charge $Q$ and a point charge $q$ a certain distance away, what is the force on the point charge?

The solution is a simple application of the method of images, but in Griffiths part of the argument relies on saying $$Q_\text{conductor} = \sum_i q_{\ \text{image}}$$ Or, that the image charges in the conductor add to the total charge. I don't really see how this is a mathematically valid statement. I was thinking it may come from the Neumann boundary conditions and integrating Poisson's eq, but I am not sure.

For reference consider Griffiths Problem 3.8 which deals with the $Q=0$ case.

An explanation would really be helpful.

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    $\begingroup$ You may want to ask a specific question here, and provide a link to "Griffith". As it stands, it's not really answerable. $\endgroup$
    – Sklivvz
    Dec 9, 2012 at 11:35
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    $\begingroup$ Basically I want to know why charge conservation is applicable for image charges. If the conductor starts with charge Q, what is to say that the image charges add to Q? $\endgroup$
    – KF Gauss
    Dec 9, 2012 at 11:51

2 Answers 2


The formula for $Q_{conductor}$ that you reference comes from the mid-way point of an argument from superposition:

1) Start by assuming the conducting sphere is grounded. (i.e. forget about the charge $Q$ for now.)

a) Use the standard method of images to replace the grounded sphere with an equivalent image charge $q_{image} = -q (a/r)$ at position $a^2/r$, where a is the radius of the conducting sphere and $r$ is the distance of charge $q$ from the center of the sphere. You can now find the field.

b) Return to the grounded sphere. The (now known) field requires charge on the surface of the sphere to terminate it. By Gauss' law, the total charge on the surface, $Q_{conductor}=q_{image}$. That's your formula (which is generalized to the case of multiple image charges).

c) You can now break the connection grounding the sphere. Nothing changes; the sphere surface remains an equipotential at ground level.

2) With the sphere no longer grounded, you can now add charge $Q-q_{image}$ to the sphere to achieve the original problem specification. Since the electrostatic forces were balanced in part 1), the new charge distributes itself uniformly over the sphere, and so acts like a point charge at the center of the sphere.

You can now add the two fields from parts 1) and 2) to find the force on charge $q$.


You may prove that the required mirror charge is the opposite of the external charge – and if you have many external charges, the solution is given by the superposition principle so the total sum of "real" external charges is equal to the sum of the mirror charges – by calculating an integral under "integrating over all angles" in Wikipedia.

Instead of reproducing the proof over there, let me offer you a more conceptual proof. Electrodynamics is "conformally invariant". It really boils down to the fact that the action $-(1/4)\int d^4 x\, F_{\mu\nu}F^{\mu\nu}$ is scale-invariant and scale invariance together with the Poincaré invariance typically implies the full conformal symmetry.

Now, the required mirror charge is clearly $-q_{\rm external}$ if the conducting surface is planar, by a $Z_2$ symmetry (which is the only way to guarantee that $\vec E$ will be orthogonal to the plane, the boundary of the conductor, which is required by the constancy of the potential in the conductor). One may generalize this statement to an arbitrary sphere by making a spherical inversion. Consider a conductor whose surface is a plane not crossing the origin $\vec x=0$. Now, perform a spherical inversion $$ (r,\theta,\phi)\to (1/r,\theta,\phi) $$ in spherical coordinates. This transformation – which will change the plane outside the origin (and going to infinity) to a (compact, not touching the infinity) sphere touching the origin – may be easily shown to be an angle-preserving, conformal transformation (essentially because it's true in 2D, because $z\to 1/z$ is a holomorphic function of a complex variable except for the pole at $z=0$) so if all the fields are transformed properly and if they solved the equations before, they will solve it afterwords, too.

But the integral $\nabla\cdot \vec E$ which is proportional to the charge in a given region is invariant under the conformal transformations because the integrand is the 2nd derivative of the potentials whose "mass dimension" is one (just like for a derivative, too). The integrand is therefore "mass cubed" which cancels against the three-dimensional integration measure. So if you apply the spherical inversion on the planar problem, you get a sphere with an external charge and a mirror charge and the charges $+q,-q$ will be exactly like before (like for the planar problem).

One would need some maths beyond the "mass dimensions" to prove that the charge is really conformally invariant but it's true.

So while the derivation needs either some abstract group theory – spherical inversion, conformal symmetry etc. – or boring integrals, the answer is Yes, the charges are the same. By the superposition principle, you may take the external charges to be any distribution. These sources may be viewed as the superposition of point-like charges, and the resulting fields $\vec E$ will be superpositions with the same coefficients, excited by the same combination of the mirror charges. So the equality between the external and mirror "total charge" will remain valid even if you take any configuration of spherical or planar conductors (planar conductors are just the $R\to\infty$ limits of the spherical ones) and any configuration of charges.

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    $\begingroup$ I don't think $q_{image}=-q_{external}$ in this geometry. Consider the case where the external charge is very far away compared to the sphere's radius: it won't take much image charge to balance the external charge's field. $\endgroup$
    – Art Brown
    Dec 9, 2012 at 22:05
  • $\begingroup$ Oops, your point seems to be right. I had to misinterpret what $Q_{conductor}$ meant, right? $\endgroup$ Dec 10, 2012 at 8:26
  • $\begingroup$ I'm actually not sure; your answer is well above my level of understanding. (Color me jealous!) The question is a bit difficult to understand; I think I finally worked it out. $\endgroup$
    – Art Brown
    Dec 10, 2012 at 8:46

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