a) An infinite plane carries charge s per unit area. Determine its electric field. b) Suppose that the plane has a straight, very narrow, infinitely-long gap of width d. What is the electric field at...

Test 1: EM – 18/02/2017 – Solutions
Each bullet point is one step in the logical chain of the argument.
1 a)
• Consider a certain amount of charge Q within a volume V bounded by a closed surface S. If
there is a change in Q with time, this change must be associated with a flow of charge through
the bounding surface. Let I be the rate of flow of charge, i.e. the cu
ent, outward through
S. This gives us the integral version of the continuity equation:
dQ
dt
= −I (1)
• We can write Q =
∫∫∫
V
ρ dV , and I =
∫∫
S
j · dS. So, dQ
dt
= d
dt
∫∫∫
V
ρ dV =
∫∫∫
V
∂ρ
∂t
dV , and,
using Gauss’s divergence theorem,
∫∫
S
j · dS =
∫∫∫
v
∇ · j dV . Thus∫∫∫
V
(∂ρ
∂t
+ ∇ · j
)
= 0 (2)
• Since this is true for every volume V , however small, we must have
∂ρ
∂t
+ ∇ · j = 0 (3)
which is the continuity equation in differential form.
—————————————————————————————-
1 b)
• In a conductor j = σE. Putting this into the continuity equation we get ∂ρ
∂t
+ ∇ · (σE) = 0.
Since σ is constant, and since ∇ · E = ρ/�o, we have
∂ρ
∂t
+
σ
�o
ρ = 0 (4)
To find ρ(r, t) we need to integrate this equation. Since we have a partial derivative here, we
need to be a little careful. When integrating an equation like dy
dx
= g(x) we get a constant of
integration, i.e. a quantity that, when is acted upon by d
dx
, yields zero. By analogy, when we
have a partial derivative, we need, in place of the usual constant of integration, a quantity that
yields zero when it is acted upon the partial derivative operator. Since ρ(r, t) is a function
of both position and time, this quantity is a function of position. So, when integrating the
equation above we get
ln ρ = − σ
�o
t+ ln f(r) (5)
where f(r) is some function of r. Here, putting t = 0, we see that f(r) is just the initial
charge density ρ(r, 0); thus
ρ(r, t) = ρ(r, 0) e−σ t/�o (6)
1
• The physical interpretation of this equation is that if there is an initial charge distribution
inside a conductor, it vanishes within a few times �o/σ, which is the natural time-scale of this
process. Since σ is the reciprocal of the resistivity, �o/σ effectively an RC time constant; in
most metals it is extraordinarily small: try determining it for copper, for example. Within
this time, all volume charge in a conductor tends to migrate to the surface. Since ∇·E = ρ/�o,
this is also the scale over which electric fields within a conductor vanish. (The movement of
charge associated with the migration of charge to the surface produces strong cu
ents; as a
esult, the magnetic fields inside a conductor can be much stronger than the electric fields.)
—————————————————————————————-
2 a)
• Symmetries :
Let the infinite,...