I've read the section in Introduction to Electrodynamics by Griffiths over and over and I still don't understand the concept of the method of images. This problem builds on the concept of using an image charge inside a spherical conducting shell (neutral).

My confusion arises from a number of things.

  1. Why do we even need to pretend to have a charge inside the conducting shell?

  2. How can I geometrically perceive the concept of potential? Currently, I imagine putting a tack on a board. Then, I loop one end of a rubber band to the tack and begin to pull the rubber band to the left. As I move farther and farther to the left, the tension in the rubber band grows to infinity. If I were to let go, the rubber band would fly towards the right with an energy related to the amount of tension in the band just before release. I imagine this when moving a charge closer to another charge. As I move one charge closer and closer to another one, the amount of energy that the charge will have once released becomes greater and greater. This is my understanding of potential. Is this correct?

  3. What does adding another charge do in the problem below?

For a grounded sphere ($V=0$) a single image charge is sufficient. But with the addition of a second image charge, the same basic model will handle the case of a sphere at any potential $V_0$ (relative to infinity). What charge should you use, and where should you put it? Find the force of attraction between a point charge $q$ and a neutral conducting sphere.

Update: I solved for the potential on the outside of the sphere and on the surface. If I'm wrong, please feel free to help me understand better.

Potential outside the sphere:

$V=\displaystyle\frac{kq}{R}$ when $r>R$.

Potential on the surface:

$V=\displaystyle\frac{kq}{R} + kq(\frac{1}{r} - \frac{1}{R})$ when $r<R$.

  • $\begingroup$ The method of images allows you to write down a solution to the boundary value problem defined by Laplace's equation + boundary conditions. Due to the uniqueness theorem, this solution is unique. However, it is in principle always possible to solve an image problem using a conventional "series solution" as explained in Chapter 3 of Griffiths (in the 4th edition, this chapter is just called "Potentials"). $\endgroup$ Commented Sep 20, 2016 at 22:39
  • $\begingroup$ We don't "pretend". The solution (if found using the series method for instance) looks exactly like that with an image inside. That's the point. The cleverness lies in the fact that you can (often, not always) reduce a lot of work by an intelligent guess, and the fact that there is a uniqueness theorem means the solution you found to Laplace's equation -- if it satisfies the boundary conditions -- is indeed the correct solution. $\endgroup$ Commented Sep 20, 2016 at 22:40
  • $\begingroup$ Your rubber band example pertains to storage of potential energy (or more precisely, elastic potential energy) in a stretched rubber band. If the elastic force is conservative (i.e. the work done by this force in stretching the band from position A to position B is independent of the "path" taken in going from A to B), then the force is described by a potential, and hence potential energy is a well defined concept. $\endgroup$ Commented Sep 20, 2016 at 22:44
  • $\begingroup$ The situation with electrostatics is a bit different. First of all, if the two charges are of the same polarity, they will repel each other more and more as they are brought closer. This will in fact correspond to greater energy of the two-charge configuration. But if the polarities are different, the force between the charges is attractive and as the charges approach each other, the energy will reduce. In either case, a potential is well defined because electrostatic electric fields are conservative: the line integral of the electric field around any closed path is zero. $\endgroup$ Commented Sep 20, 2016 at 22:46
  • $\begingroup$ Hint for the final problem you ask: forgetting about images, what is the potential outside a spherical cavity (of radius $R$) which contains a charge $q$ somewhere inside, measured at a point at a distance $r$ ($r \geq R$) from the center of the cavity? What is the potential at $r = R$ (i.e. at the surface of the sphere)? $\endgroup$ Commented Sep 20, 2016 at 22:48

2 Answers 2


You do know where $\nabla$$^2$V = -$\rho$/$\epsilon$$_0$ comes from. it comes directly from the divergence of the electric field and from the relation E = -$\nabla$V. There is a proof in Griffiths which shows why the solution of V is unique, if at all the boundaries of the system the potential is known, and the charge within the system remains the same. Given the uniqueness of this system, if one can intuitively guess a situation in which these situations hold, then ANY problem having the same initial conditions will have the same solution for potential. That is the basic idea behind image charge. To reduce the problem.


Presumably you understand how to solve the problem of finding the potential outside a conducting sphere of radius $R$ (held at $V = 0$) when a point charge $q$ is placed at some distance $a$ outside the sphere ($a > R$), using the method of images (placing an image charge $q'$). To be sure, this is Example 3.2 in the textbook by Griffiths.

The problem you have posed asks you what you must do to solve this problem when the conducting sphere is held at a potential $V = V_0$. The answer is that you place a point charge $q''$ at the center: doing so keeps the sphere an equipotential, but just raises the potential at the surface to $V_0$ if

$$q'' = \frac{V_0 R}{k}$$

($k = 1/(4\pi\epsilon_0)$ in the notation you used in the comments.)

For the sphere to stay neutral, the two image charges $q'$ and $q''$ must add up to zero:

$$q' + q'' = 0$$

Then the force on external charge $q$ (which is the physical charge, in the region where you're effectively solving Laplace equation if you like) is due to $q'$ and $q''$ and is given by

$$F = kq\left[\frac{q''}{a^2} + \frac{q'}{\left(a - (R^2/a)\right)^2}\right]$$

Note that what you have done is solve Laplace equation in the region outside the spherical cavity (where the physical charge $q$ is placed). You cannot ask (in the above problem) what the potential or field inside the cavity are.

What you asked in the comments was therefore a different problem: consider a grounded conducing spherical cavity of radius $R$, with a charge $q$ placed somewhere inside it. What is the electric field and potential inside? Note that this is a different problem and should not be confused with the image problem above because for this problem you will effectively solve the Laplace equation inside the sphere. For more on this problem, see Problem 2.2 in http://users-phys.au.dk/thostrup/pdf/chapter2.pdf.

I will edit this answer to resolve any outstanding doubts/concerns you might have.

  • $\begingroup$ Why can't I ask what the field and the potential are inside the sphere? Is it because there are only ''fake'' charges inside? $\endgroup$ Commented Sep 21, 2016 at 0:32
  • $\begingroup$ Yes, that's right. Using the method of images, you cannot find what the potential inside is, as you're solving the equation in the complement of the space where the image is. You can argue that one can write the most general solution to the Laplace equation inside the sphere and outside the sphere, match them at the surface (since $V$ is continuous) and impose the condition that the tangential component of the electric field at the conductor's surface vanishes. This would determine $V(r)$ for both regions. See the comment in Griffiths's book at the bottom of Example 3.2 on page 128. $\endgroup$ Commented Sep 21, 2016 at 0:41
  • $\begingroup$ And see section 3.3 of Griffiths for a variety of problems where you solve Laplace equation in different regions through a series solution and patch up the descriptions in different regions using boundary conditions. This method (although more general) is more tedious but is often the only way out when writing the Green function for the arrangement isn't obvious from the method of images. $\endgroup$ Commented Sep 21, 2016 at 0:43
  • $\begingroup$ Also, what does it mean to say that the sphere is ''held at a potential $V_0$? Does it mean that you've placed it in a specific location where the potential is a certain value $V_0$? $\endgroup$ Commented Sep 21, 2016 at 0:43
  • $\begingroup$ Holding the sphere at a certain value $V_0$ means that the (conducting) sphere has been connected to a battery that keeps at a fixed potential $V_0$. This has nothing to do with where the sphere is: it could be anywhere. This just says that $V(r = R) = V_0$ always. It is a boundary condition on the potential on the surface. The caveat is that the source/sink (battery) which maintains this must source or sink any amount of charge to maintain the equipotential nature of the surface of the sphere. $\endgroup$ Commented Sep 21, 2016 at 0:46

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