Uniformly Charged Finite Plane

To find E above the midpoint of a plane of length L , width W and
uniform surfact charge density σ (charge per area), slice the plane
into many thin rods.
z
θ θ
dE
dE
r
z θ
x
L
x
dq
W
dq = σ dA = σ L dx

The horizontal components of dE from two symmetrically located
thin slices will cancel and the z-components sum, so Enet is along
the z-axis.

dEz = dEslice cosθ ; Ez = 2
z 
dEslice
∫
r
half plane
We can find the the electric field due to a thin slice by using our
earlier result for the electric field along the perpendicular bisector of
a uniformely charged thin rod of length L and charge Q .
Here, the thin rod has charge dq = σ L dx and the point on the z-axis
is a perpendicular distance r away from the rod:
E
h

E=

2ke σ L dx
dEslice =
r 4 r2 + L2
2ke Q
h 4 h2 + L2
L, Q

Substituting for dEslice
W /2
2ke σ L dx
dx
=
4k
σ
Lz
e
∫0 x 2 + z2 4 x 2 + z 2 + L2
r r 4 r 2 + L2
W /2 z
Ez = 2 ∫ 0
(
) (
)
We can evaluate the integral (see “Doing the integral”) to find the
simple expresion:
Efinite plane = Ez = 4keσ arctan
L /W
2 ( z / W ) 4 (z /W ) + ( L /W ) +1
2
2
As z → 0 , arctan ∞ returns a value of π / 2 and Efinite plane → E0 = 2π kσ
This is the same result we had for the infinitely large circular disk.
Dividing both sides by E0
Efinite plane 2
= arctan
E0
π
L /W
2 ( z / W ) 4 (z /W ) + ( L /W ) +1
2
2
For a given plane, L /W is known, so Ez / E0 is simply a function of
z /W , the relative height above the plane.