Ex. Electric Field due to a Finite Charged Rod
Find the electric field some distance y above a uniformly charged finite rod
E
dE2
θ
θ
θ
θ
dqi represents the amount of
charge on the rod piece of length
dxi. The corresponding electric
field at a point P due to dqi is dEi.
dE1
P
r =x +y
2
r
r
y
2
cos θ =
dq2
dq1
x
dx1
x
sin θ =
dx2
y
2
=
r
x
y
x +y
2
=
r
2
x
x +y
2
2
l
l
The total length of the rod is L (L = 2l).
Since this is a uniform charged rod Æ
λ=
dq
dqi = λ dxi
Î
dl
At point P:
dEi =
ke dqi
r
2
rˆ =
ke dqi
r
2
( sin θ xˆ + cosθ yˆ ) = ⎛⎜
ke dqi
⎝ r
2
⎞
⎠
⎛ ke dqi cosθ ⎞ yˆ
⎟
⎝ r2
⎠
sin θ ⎟ xˆ + ⎜
The total electric field E due to dqi can be found by summing up (integrating) all the dqi
contributions for the entire length of the rod:
l
E=
⎛ ke dq sin θ ⎞ xˆ + ⎛ ke dq cos θ ⎞ yˆ
⎟ ⎜ 2
⎟
∫ ⎜ r2
⎠ ⎝ r
⎠
−l ⎝
Substituting for dq, r, sin θ and cos θ and separating the integral into its two components we get:
⎛ k xλ dx
e
E = ∫⎜
2
2
⎜
−l ⎝ ( x + y )
l
3
2
l ⎛
⎞
k yλ dx ⎞
⎟ yˆ
⎟ xˆ + ∫ ⎜ e
⎟ −l ⎜ ( x 2 + y 2 ) 3 2 ⎟
⎠
⎝
⎠
The first integral ( x̂ ) direction, is an odd function [ f ( − x ) = − f ( x ) ] . Odd functions evaluated over
symmetric limits always integrate to 0.
⎛ k yλ dx ⎞
e
ˆ
E = 0+ ∫⎜
3 ⎟y
2 ⎟
2
2
⎜
+
x
y
) ⎠
−l ⎝ (
l
Î
Rearranging the limits of the integral:
⎛ k yλ dx ⎞
∫0 ⎜ ( x 2e + y 2 ) 3 2 ⎟⎟ yˆ
⎝
⎠
l
E=2 ⎜
Integrating yields:
E = 2ke λ y
E=
⎡
⎢
⎣y
l
x
2
2ke λl
y y +l
2
2
2
y +x
2
⎤
⎥ yˆ
⎦0
yˆ
This implies that the electric field is always perpendicular to the center of the wire, since there is
no x component in the final expression for E.