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Chapter 20
Problem 56
y
†
r
y
L
x
Solution
For a uniformly charged rod with charge, Q, and length, L; show that the electric field at a distance, x,
along the direction of the rod is E = kQ/[x(x + L)].
Begin with the formula for electric field with continuous media.
Z
kdq
r̂
E=
r2
Since the charge is uniformly distributed over the length of the rod, the charge on an infinitesimal slice of
the rod is
dq = λdy =
Q
dy
L
The point of interest is a distance of L + x away from the origin. Therefore, the distance between the
infinitesimal slice and the point of interest is
r =L+x−y
The direction vector r̂ is in the positive x direction. We are integrating over all of the charge on the rod;
therefore, the limits of integration are y = 0 to y = L. Therefore, the integral becomes
Z
Z L
Qdy
λdy
E=k
r̂
=
k
î
2
r2
0 L(L + x − y)
To solve this use u-substitution. Let p = L + x − y. Then dp = −dy and
y=L
Z y=L
−Qdp
kQ 1 E=k
î =
î
2
L
p y=0
y=0 L(p)
Back substituting for p gives
y=L
1
1
1
kQ
kQ
E=
î
=
î
−
L
L + x − y y=0
L
L+x−L L+x−0
kQ 1
1
E=
î
−
L
x L+x
Get a common denominator
kQ
L+x
x
kQ L + x − x
E=
î
−
=
î
L
x(L + x) x(L + x)
L
x(L + x)
kQ
L
kQ
E=
î
=
î
L
x(L + x)
x(x + L)
If this process were repeated for the left side of the rod, the setup of the integral would be different, but
the resultant magnitude would be the same. The direction vector would be in the −î direction.
†
Problem from Essential University Physics, Wolfson
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