PHYS 208
Spring 2009
#2
1
Electric Field
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
Q
Q
Rdθ =
dθ.
2πR
2π
dq = λ ds =
The distance between dq and P is r, which is
r=
√
x2 + R 2 ,
and, from the figure,
cos α =
x
x
=√
,
r
x 2 + a2
sin α =
R
R
=√
.
r
x2 + R 2
Note that both r and the angle α are identical for any point ds on the circle so that they do not depend on the integration variable θ. From the geometry of the system, we can see that the resulting
~ lies on the yz plane and
electric field points the positive x-direction. (The other component of dE
is canceled out.) Then
~ = dE cos α îx ,
dE
where
dE =
1
dq
.
4π0 x2 + R2
1
Since
Z
Ex =
~ x,
dE
we get
Ex =
=
=
=
Z 2π
dq
1
cos α
2
4π0 0 x + R2
Z 2π
1
Q
1
x
√
dθ 2
2
4π0 0 2π x + R x2 + R2
Z 2π
x
1 Q
dθ
4π0 2π (x2 + R2 )3/2 0
1
Qx
.
4π0 (x2 + R2 )3/2
Therefore,
Qx
1
îx .
4π0 (x2 + R2 )3/2
Therefore, when P is at the center of the ring, E = 0. When P is much farther from the ring
(x R), we get
~ =
E
Ex =
Qx
1 Qx
1 Q
1
≈
=
,
2
2
3/2
3
4π0 (x + R )
4π0 x
4π0 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
The circumference of the ring Ris 2πr and the length in the radial direction is dr, so the area is
R
~ due to the ring is pointing
2πrdr. You should verify that 0 dq = Q. Now, the electric field dE
the positive x-direction and its magnitude is
dEx =
dqx
1
,
2
4π0 (x + r2 )3/2
from the previous example. Thus,
Z R
1
dqx
dEx =
2
4π0 0 (x + r2 )3/2
r=0
Z R
1
σ2πrdrx
4π0 0 (x2 + r2 )3/2
Z
rdr
σx R
2
20 0 (x + r2 )3/2
R
σx
1
−√
20
x2 + r 2 0
1
1
σx
+
−√
20
x2 + R 2 x
σ
x
.
1− √
20
x2 + R 2
Z
Ex =
=
=
=
=
=
r=R
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
−1/2
x
x
R2
1 R2
R2
√
p
1−
=1−
=1− 1+ 2
=1− 1−
= 2.
x
2 x2
2x
x2 + R 2
x 1 + R2 /x2
Therefore,
Ex ≈
σ R2
Q R2
1 Q
=
=
,
2
2
2
20 2x
πR 20 2x
4π0 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,
x
x
1− √
≈ 1 − ≈ 1,
R
x2 + R 2
therefore,
σ
Ex =
.
20
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
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/(20 ). 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
E = 0,
σ
pointing upward,
Region II, between the planes
E=
0
Region III, below the lower plane
E = 0.
Region I, above the upper plane
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