Week 5 Lex 2: The Method of Images.

The method of images for electrostatic problems is a way to obtain the potential V for a problem by identifying it with the known ( simpler to calculate) potential V of a different electrostatic problem but satisfying the same boundary condition as the original V in the region of interest. Uniqueness theorem then ensures that the two potentials are identical in the region of interest.


Classic Image Problem I:


A point Charge q placed at a distnace d in front of an infinite conducting plane which is grounded, to find the potential at a point above the plane.


The trick to solve this problem is to identify that one of the equipotential at the mid point between two equal and opposite charge +q and -q is a plane and its at V=0. This is identical to the case of a point charge in front of a conducting plane at ZERO potential because the potential satisfies the same boundary conditions V=0 at z=0 taking the z-axis to be perpendicular to the plane.


Thus the image problem is that of two charges +q and -q situated at equal distance from the origin in the +ve and -ve direction of the z-axis. Now the potential due to this arrangement is very simple to calculate by superposition of the potential due to each of the two charges the image charge and the point charge.


Note that the solution is only valid above the conducting plane. Below the plane the situation is completely different but that is excluded from our region of interest. Uniqueness theorem ensures that in the region of interest this is the only UNIQUE solution that satisfies the given boundary conditions. Hence the solution for the potential


Note also that the image charge cannot be placed in the region of interest because that would completely change the original problem we are seeking a solution to. Its important to understand that the method of images is a method to find an alternate simple problem whose potential is identical with that of the original problem due to both satisfying the same boundary conditions in the region of interest.


The solution can be easily extended to that of the isolated  ( ungrounded ) conducting plane.


Classic Image Problem II:


A point charge q is placed outside at the distance a from the center of a conducting sphere of radius Rwhich is grounded, to find the potential everywhere outside. The trick for this is to identify that one of the equipotentials for a system of two unequal charges q and q' is a sphere. So the only thing now is to set the boundary conditions.
Since the original sphere is grounded ( at ZERO potential) in the new problem we will have to adjust q' so that the spherical surface is at V=0. Now its esay to calculate V ( IPSA Problem) for the two unequal charges and see that the image charge must be set at x=R²/a from the center and its magnitude should be q'=-R/a x q to make the potential V(R)=0.


The problem with the grounded sphere may easily be extended to cover an isolated ( ungrounded) sphere. The sphere would be an equipotential now at V=V₀


It is now possible to use these two image problems and superposition of potentials to solve more complex problems which I have mentioned in the lecture and some of which will be discussed in the future tutorials.