Download Force on a surface charge distribution

W3007 – Electricity and Magnetism
1
Force on a surface charge distribution
Consider a generic smooth surface, with a surface charge density σ(r), in the presence of an electric
field. What is the force acting on a infinitesimal area element da? It is
dq = E
σ da .
dF = E
(1)
This expression is ambigous, for the electric field is discontinuous right at the surface. Should we
use the field right above or right below the surface? To answer this question, we have to give up
the idealization of having an infinitely thin layer of charges, and consider the more realistic case
in which charges are distributed with some volume density ρ(r) in some thin layer of thickness .
If we zoom in very close to the area element da we are interested in, the surface looks flat (any
smooth surface looks flat if you look close enough!) and the charge distribution looks like a flat
slab of charges. In this approximation, let’s choose coordinates such that z is perpendicular to
the surface, and x and y run parallel to it. The charge layer extends from z = 0 to z = ε:
z
E above
ε
ρ
x
0
E below
Now, the crucial fact we will make use of, is that the z-dependence of the charge density and
of the electric file is quite abrupt, whereas the x- and y-dependences are smooth. This is because
in the limit in which the charge layer is infinitely thin (ε → 0), the electric field and the density
are discontinuous in crossing the surface, i.e. in moving along z across z = 0. This means that we
and ρ are functions of z alone. We will not assume anything else about this
can assume that E
functions—the charge can be distributed in any way inside the layer. We will of course assume
Maxwell’s equations, and the boundary conditions:
above
E(z)
=E
below
E(z)
=E
for z > ε
(2)
for z < 0
(3)
and
ρ(z) = 0
for z < 0 or z > ε .
(4)
Also, since it is the z-component of the electric field that is ambiguous—eq. (1) makes perfect sense
because those are continuous at the surface—we will concentrate
for the parallel components of E,
2
Force on a surface charge distribution
just on that. So, the force per unit area acting on an infinitesimal volume element dτ = dadz is
dFz = Ez (z) dq = Ez (z) ρ(z) dadz .
To get the total force per unit area, we divide by da, and integrate along z:
ε
dFz
=
Ez (z) ρ(z) dz
da
0
(5)
(6)
Now we make use of Gauss’s law,
·E
= d Ez (z) = 1 ρ(z)
∇
dz
0
(7)
and rewrite the integral as
ε
dFz
d
Ez Ez dz
= 0
da
dz
0 ε
d 2
0
E dz
=
2 0 dz z
0 2
Ez above − Ez2 below
=
2
0 =
Ez above − Ez below Ez above + Ez below
2
(8)
(9)
(10)
(11)
The first parenthesis is precisely the discontinuity of the electric field at the surface, which combines
with 0 to yield σ. The second parenthesis combines with the factor of 1/2 to yield the average
electric field:
dFz
= σEz average .
(12)
da
are continuous at the surface, and the average between their
Since the parallel components of E
value above and their value below is equal to either, we can rewrite the whole force (per unit area)
vector as
dF
average ,
= σE
(13)
da
where
below ) .
above + E
average ≡ 1 E
(14)
E
2