70 Week 2: Continuous Charge and Gauss’s Law
so
dx=
y dθ
cos^2 θ
(55)
and
r=
y
cosθ
(56)
If we substitute these into the expressions above we get:
dEy=
keλ dx
r^2
cosθ=keλ
(
y dθ
cos^2 θ
)(
cos^2 θ
y^2
)
cosθ=
keλ
y
cosθdθ (57)
which lookseasyto integrate! The limits of integration are the angles to the dotted lines that point
at the ends of the line, which we will callθ 1 on the left,theta 2 on the right. Thus:
Ey=
keλ
y
∫θ 2
θ 1
cosθdθ=
keλ
y
(sinθ 2 −sinθ 1 ) (58)
(where we should carefully note thatθ 1 in the figure above isnegativeas drawn).
If we evaluateExeverything is the sameexcept that there is an overall minus sign and we
integrate over sinθ dθinstead, to get:
Ex=−keλ
y
∫θ 2
θ 1
sinθdθ=keλ
y
(cosθ 2 −cosθ 1 ) (59)
An interesting consequence of this result is that we can easily evaluate the field a distancey
away from aninfiniteline of charge (that still has a uniform charge per unit lengthλ. In that case,
θ 1 =−π/2 andθ 2 =π/2. We get:
Ex(∞) = 0 (60)
Ey(∞) =
2 keλ
y
(61)
where we should recall thateverypointPhas anx-coordinate in the middle of an infinite line of
charge! Remember this result for later, where we will obtain it again using Gauss’s Law.
Example 2.1.3: Circular Disk of Charge
dE dE
z (z + r )^2 2 1/2
θ
φ
θ r
x
y
z
z
dr
r d dA = r dr d θ
σ
R
P
Figure 12: A charged disk with charge per unit areaσ.
In figure 12 above we see a disk of charge with a uniform charge density:
σ=
Q
πR^2