Jackson 2.28 Homework Problem Solution
Dr. Christopher S. Baird
University of Massachusetts Lowell
PROBLEM:
A closed volume is bounded by conducting surfaces that are the n sides of a regular polyhedron
(n = 4, 6, 8, 12, 20). The n surfaces are at different potentials Vi, i = 1, 2, …, n. Prove in the simplest
way you can that the potential at the center of the polyhedron is the average potential on the n sides.
This problem bears on Problem 2.23b, and has an interesting similarity to the result of Problem 1.10.
SOLUTION:
Because the potential obeys the superposition principle, we can break down the total potential at the
center point in the polyhedron due to the n sides at different potentials Vi as a sum of potentials, each
one due to one side being held at its potential Vi and all other sides being held at zero:
Φ( V1, V2, V3,. .. , V n)= Φ(V1, 0, 0,. .. ,0)+Φ(0,V2, 0,... ,0)+...+Φ (0, 0, 0,. .. , V n)
Φ( V1, V2, V3,. .. , V n)=∑
i=1
n
Φ(ith side at Vi, all others at 0)
Φ( V1, V2, V3,. .. , V n)=∑
i=1
n
Φi where Φi=Φi(ith side at Vi, all others at 0)
Now consider a special case where all the walls are held at the same potential Vi = V. In this case, the
relaxation method tells us that the entire interior, including the center point must also be at this constant
potential so that
Φ=V
. Because the center point is equally distant from each face and they all have the
same area, the potentials at the center due to each face are all equal if all the faces are equal:
Φi=Φ0
.
Plugging these values into the last equation above, for this special case, we have:
V=∑
i=1
n
Φ0
V=nΦ0
Φ0=V
n
The contribution at the center from each face is therefore V/n where V is the potential of the face. Now,
if a face a held at a potential of zero, it obviously contributes nothing in the sense of the superposition
of potentials. (Surfaces at zero potential obviously contribute to a problem overall because they provide
boundary conditions that must be met. A surface at zero potential is very different from no surface.)
This means that if all other faces are held at a potential of zero except face i which is held at potential
Vi, the potential at the center will be Vi/n by the equation above:
