The document summarizes the electric field calculations for three scenarios: 1) A ring of uniformly distributed charge - the electric field decreases with the square of the distance outside the ring and is zero at the center. 2) A uniformly charged disk - the electric field decreases with the inverse of the distance outside the disk and is uniform across an infinite sheet. 3) Two oppositely charged infinite sheets - the electric field is zero outside the sheets and uniform between them, pointing from the positively charged sheet.

1. PHYS 208 Spring 2009
#2 Electric Field
1 Electric field of a ring of charge
A ring-shaped conductor with radius R carries a total charge Q uniformly distributed around it.
Find the electric field at a point P that lies on the axis of the ring at a distance x from its center.
(The ring lies on the yz plane.)
We first calculate the linear charge density,
λ =
Q
2πR
.
Then dq is
dq = λ ds =
Q
2πR
Rdθ =
Q
2π
dθ.
The distance between dq and P is r, which is
r =
√
x2 + R2,
and, from the figure,
cos α =
x
r
=
x
√
x2 + a2
, sin α =
R
r
=
R
√
x2 + R2
.
Note that both r and the angle α are identical for any point ds on the circle so that they do not de-
pend on the integration variable θ. From the geometry of the system, we can see that the resulting
electric field points the positive x-direction. (The other component of dE lies on the yz plane and
is canceled out.) Then
dE = dE cos αˆix,
where
dE =
1
4π 0
dq
x2 + R2
.
1

2. Since
Ex = dEx,
we get
Ex =
1
4π 0
2π
0
dq
x2 + R2
cos α
=
1
4π 0
2π
0
Q
2π
dθ
1
x2 + R2
x
√
x2 + R2
=
1
4π 0
Q
2π
x
(x2 + R2)3/2
2π
0
dθ
=
1
4π 0
Qx
(x2 + R2)3/2
.
Therefore,
E =
1
4π 0
Qx
(x2 + R2)3/2
ˆix.
Therefore, when P is at the center of the ring, E = 0. When P is much farther from the ring
(x R), we get
Ex =
1
4π 0
Qx
(x2 + R2)3/2
≈
1
4π 0
Qx
x3
=
1
4π 0
Q
x2
,
which means that the ring would appear like a point charge of Q.
2 Electric field of a uniformly charged disk
Find the electric field caused by a disk of radius R with a uniform positive surface charge density
σ at a point along the axis of disk a distance x from its center. Assume that x is positive.
To solve this problem, we use the result of the previous example, i.e. the electric field caused
by a ring of charge. First, consider the ring of width dr. Then the integration variable is r and is
from r = 0 to r = R. Since the surface charge density is σ = Q/(πR2
), the charge dq carried by
the ring is
dq = σ2πrdr.
2

3. The circumference of the ring is 2πr and the length in the radial direction is dr, so the area is
2πrdr. You should verify that
R
0
dq = Q. Now, the electric field dE due to the ring is pointing
the positive x-direction and its magnitude is
dEx =
1
4π 0
dqx
(x2 + r2)3/2
,
from the previous example. Thus,
Ex =
r=R
r=0
dEx =
1
4π 0
R
0
dqx
(x2 + r2)3/2
=
1
4π 0
R
0
σ2πrdrx
(x2 + r2)3/2
=
σx
2 0
R
0
rdr
(x2 + r2)3/2
=
σx
2 0
−
1
√
x2 + r2
R
0
=
σx
2 0
−
1
√
x2 + R2
+
1
x
=
σ
2 0
1 −
x
√
x2 + R2
.
Again, we can verify that the disk would look like a point charge when x is much larger than
R. To see this, use the Taylor expansion
(1 + a)n
≈ 1 + na, when a 1.
Then, if x R, R/x 1, and
1 −
x
√
x2 + R2
= 1 −
x
x 1 + R2/x2
= 1 − 1 +
R2
x2
−1/2
= 1 − 1 −
1
2
R2
x2
=
R2
2x2
.
Therefore,
Ex ≈
σ
2 0
R2
2x2
=
Q
πR22 0
R2
2x2
=
1
4π 0
Q
x2
,
where we have used σ = Q/(πR2
).
Another interesting limit is when R x, i.e., the electric field produced by an infinite sheet
of charge. In this case,
1 −
x
√
x2 + R2
≈ 1 −
x
R
≈ 1,
therefore,
Ex =
σ
2 0
.
This shows that the electric field produced by an infinite sheet of charge takes the same value
independent of the position x, so is a constant electric field.
3

4. 3 Electric field of two oppositely charged infinite sheets
Two infinite plane sheets are placed parallel to each other, separated by a distance d. They have
uniform charge distribution and the upper plane has surface charge density −σ and the lower plane
has +σ. Find the electric fields in the three regions shown in the figure.
From the previous example, we know that the electric field produced by an infinite plane of
surface charge density is E = σ/(2 0). The lower plane has negative charge so the electric field
is toward the plane and the upper plane has positive charge so the electric field is from the upper
plane. Since these electric field is constant we have electric fields as shown in the figure. Therefore,
in Region I, the two electric field is antiparallel with the same magnitude, so there are canceled
out. So we have
Region I, above the upper plane E = 0,
Region II, between the planes E =
σ
0
pointing upward,
Region III, below the lower plane E = 0.
4