Assume that we have positively charged two solid conducting spheres of the same radius and material. It is evident that the surface charge is uniform on them to eliminate the internal electric field everywhere inside the spheres. However, how does the charge distribution change if we bring them close to each other? Recall that we place them close (not in contact) to each other. Does the exerted Coulomb's force upset the uniformity of the surface charges on both spheres? How can we find the new charge distribution so that the internal electric fields remain zero?

  • $\begingroup$ @AXensen Yes, they do. $\endgroup$ Commented Sep 23, 2023 at 8:23
  • $\begingroup$ I'd say, yes they change and you can find the charge distribution solving a boundary element problem, if you know the potential of the 2 spheres. Try to have a look here physics.stackexchange.com/q/781452. I'll have a think if there is some simple way to exploit symmetry to get the solution. The internal field = 0 appears as a boundary condition of the differential equation $\endgroup$
    – basics
    Commented Sep 23, 2023 at 8:26
  • $\begingroup$ Electrostatics of two charged conducting spheres $\endgroup$
    – Farcher
    Commented Sep 23, 2023 at 8:55
  • $\begingroup$ @Farcher Seems to be a useful article, thanks. $\endgroup$ Commented Sep 23, 2023 at 8:59

1 Answer 1


Some of your question can be answered conceptually. A uniformly charged sphere on its own is an equipotential surface, and it makes a potential outside of itself of $(1/4\pi\epsilon_0)Q/r$. Thus if two uniformly charged spheres are brought near eachother, they will no longer be equipotentials. The electric field created by sphere $1$ makes sphere $2$ no longer an equipotential. If the spheres are conducting, the charges on sphere $2$ will move to counteract the influence of sphere $1$ (and, of course, vice versa).

One way of calculating this is with a multipole expansion of the potential of the spheres. If the spheres are far apart, then to leading order sphere $1$ creates a constant electric field at sphere $2$. This can be fixed by giving each sphere a dipole moment (a charge distribution $\propto-\cos\theta$, where $\theta$ is the angle formed by the center of the sphere, the point on the sphere in question, and the other sphere). Such a charge distribution's magnitude can be adjusted to cancel the electric field within the conductor. Of course, higher order multipole moments can be introduced for more precision.

However, this particular problem can also be solved exactly with the method of images. We'll construct an image charge solution to the electric field outside of the conducting spheres. Our solution needs to satisfy three requirements in order to be the uniquely correct electric field throughout space:

  1. The total charge on each sphere is $Q$. This means all image charges inside one sphere need to add to $Q$.
  2. The spheres are equipotentials. This means for each image charge we place outside of a sphere, it should have a partner inside the sphere whose magnitude and position is dictated by the spherical image charge solution. A single charge can be placed at the center of the sphere whose magnitude can be adjusted to achieve condition (1). It does not need an image charge outside because it does not disrupt the equipotential on the surface.
  3. It needs to satisfy the Laplace equation, which is to say that all image charges must be located within the spheres.

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We solve this problem with an infinite series of image charges. First, we place initial charges $Q$ in the center of each sphere. Each charge gets an image charge in the other sphere. Google "spherical image charge solution" if you don't know where to place those image charges or how big to make them. Then those image charges get image charges... etc. This converges because each image charge gets smaller and smaller by roughly a factor of $r/d$. When you've listed out all the image charges, sum up the total charge, and rescale to get the desired total charge.

Then you can calculate the distribution of real charges on the surface of the spheres. Just find the electric field at a given point on the surface of the sphere from the image charges and multiply by $\epsilon_0$ to get the surface charge density (of real charges) at that point.


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