10Gauss' Law and Poisson's equation
IA Vector Calculus
10.2 Laws of electrostatics
Consider a distribution of electric charge at rest. They produce a force on a
charge q, at rest at r, which is proportional to q.
Definition
(Electric field)
.
The force produced by electric charges on another
charge q is F = qE(r), where E(r) is the electric field, or force per unit charge.
Again, this is conservative. So
I
C
E · dr = 0
for any closed curve C. It also obeys
Law (Gauss’ law for electrostatic forces).
Z
S
E · dS =
Q
ε
0
,
where ε
0
is the permittivity of free space, or electric constant.
Then we can write it in differential form, as in the gravitational case.
Law (Gauss’ law for electrostatic forces in differential form).
∇ · E =
ρ
ε
0
.
Assuming constant (or no) magnetic field, we have
∇ × E = 0.
So we can write E = −∇ϕ.
Definition
(Electrostatic potential)
.
If we write
E
=
−∇ϕ
, then
ϕ
is the
electrostatic potential, and
∇
2
ϕ =
ρ
ε
0
.
Example.
Take a spherically symmetric charge distribution about
O
with total
charge
Q
. Suppose all charge is contained within a radius
r
=
a
. Then similar
to the gravitational case, we have
E(r) =
Q
ˆ
r
4πε
0
r
2
,
and
ϕ(r) =
−Q
4πε
0
r
.
As
a →
0, we get point charges. From
E
, we can recover Coulomb’s law for the
force on another charge q at r:
F = qE =
qQ
ˆ
r
4πε
0
r
2
.
Example (Line charge). Consider an infinite line with uniform charge density
per unit length σ.
We use cylindrical polar coordinates:
z
r =
p
x
2
+ y
2
E
By symmetry, the field is radial, i.e.
E(r) = E(r)
ˆ
r.
Pick
S
to be a cylinder of length
L
and radius
r
. We know that the end caps do
not contribute to the flux since the field lines are perpendicular to the normal.
Also, the curved surface has area 2πrL. Then by Gauss’ law in integral form,
Z
S
E · dS = E(r)2πrL =
σL
ε
0
.
So
E(r) =
σ
2πε
0
r
ˆ
r.
Note that the field varies as 1
/r
, not 1
/r
2
. Intuitively, this is because we have
one more dimension of “stuff” compared to the point charge, so the field does
not drop as fast.