Jackson 1.6 Homework Problem Solution
Dr. Christopher S. Baird
University of Massachusetts Lowell
PROBLEM:
A simple capacitor is a device formed by two insulated conductors adjacent to each other. If equal and
opposite charges are placed on the conductors, there will be a certain difference of potential between
them. The ratio of the magnitude of the charge on one conductor to the magnitude of the potential
difference is called the capacitance (in SI units it is measured in farads). Using Gauss's law, calculate
the capacitance of
(a) two large, flat, conducting sheets of area A, separated by a small distance d;
(b) two concentric sphere with radii a, b (b > a)
(c) two concentric conducting cylinders of length L, large compared to their radii a, b (b > a).
(d) What is the inner diameter of the outer conductor in an air-filled coaxial cable whose center
conductor is a cylindrical wire of diameter 1 mm and whose capacitance is 3 x 10-11 F/m? 3 x 10-12 F/m?
SOLUTION:
In each case, the dimensions are such that we can neglect fringe effects and therefore can take
advantage of symmetry.
(a) Parallel Plates
If we place a charge of -Q on a plate at z = 0 and a charge of +Q on a plate at z = d, then a constant
voltage difference V is established between the two plates. The relationship between potential and
electric field is:
E=−∇
For a one-dimensional system (as this one is because the sheets are large enough to be approximated as
infinite), the magnitude reduces to:
∣
E
∣
=d
dz
For a uniform electric field (as this must be because of the symmetry) this becomes:
E=
z
The potential difference between the two plates is then:
