Simple Nature - Light and Matter

d/Discussion question A.

This can indeed be constant, but only ifnis 0 or−3, i.e.,pis 1

or−2. The second solution gives a divergence which is constant

andzero: this is the solution for theoutsideof the sphere! The

first solution, which has the field directly proportional tor, must

be the one that applies to the inside of the sphere, which is what

we care about right now. Equating the coefficient in front to the

one in Gauss’ law, the field is

E=

4 πkρ

3

rˆr.

The field is zero at the center, and gets stronger and stronger as

we approach the surface.

Discussion Questions

A As suggested by the figure, discuss the results you would get by

inserting the div-meter at various locations in the sine-wave field.

10.7.2 Poisson’s equation and Laplace’s equation
Gauss’s law, divE= 4πkρ, can also be stated in terms of the
potential. SinceE=∇V, we have div∇V = 4πkρ. If we work out
the combination of operators div∇in a Cartesian coordinate system,
we get∂^2 /∂x^2 +∂^2 /∂y^2 +∂^2 /∂z^2 , which is called the Laplacian and
notated∇^2. The Laplacian is discussed in more detail on p. 910.
The version of Gauss’s law written in terms of the potential,

∇^2 V = 4πkρ,

is called Poisson’s equation, while in the special case of a vacuum,
withρ= 0, we have
∇^2 V = 0,
known as Laplace’s equation. Many problems in electrostatics can
be stated in terms of finding potential that satisfies Laplace’s equa-
tion, usually with some set ofboundary conditions. For example,
if an infinite parallel-plate capacitor has plates parallel to thex-y
plane at certain given potentials, then these plates form a boundary
for the region between the plates, and Laplace’s equation has a so-
lution in this region of the formV =az+b. It’s easy to verify that
this is a solution of Laplace’s equation, since all three of the partial
derivatives vanish.

10.7.3 The method of images
A car’s radio antenna is usually in the form of a whip sticking
up above its metal roof. This is an example involving radio waves,
which are time-varying electric and magnetic fields, but a similar,
simpler electrostatic example is the following. Suppose that we po-
sition a chargeq >0 at a distance`from a conducting plane. What
is the resulting electric field? The conductor has charges that are
free to move, and due to the field of the chargeq, we will end up with
a net concentration of negative charge in the part of the plane near

Section 10.7 Gauss’ law in differential form 655