Simple Nature - Light and Matter

brute-force approach, then, would be to evaluate the integral∫ E=
|dE|cosθ, where dEis the contribution to the field from a charge
dqat some point along the rod, andθis the angle dEmakes with
the radial line.

It’s easier, however, to find the potential first, and then find the
field from the potential. Since the potential is a scalar, we simply
integrate the contribution dV from each charge dq, without even
worrying about angles and directions. Letz be the coordinate
that measures distance up and down along the rod, withz= 0 at
the center of the rod. Then the distance between a pointzon the
rod and the point of interest isr=

√

z^2 +R^2 , and we have

V=

∫

kdq

r

=kλ

∫+L/ 2

−L/ 2

dz

r

=kλ

∫+L/ 2

−L/ 2

dz

√

z^2 +R^2

The integral can be looked up in a table, or evaluated using com-
puter software:

V=kλln

(

z+

√

z^2 +R^2

)∣∣

∣

+L/ 2

−L/ 2

=kλln

(

L/2 +

√

L^2 /4 +R^2

−L/2 +

√

L^2 /4 +R^2

)

The expression inside the parentheses can be simplified a little.
Leaving out some tedious algebra, the result is

V= 2kλln

(

L

2 R

+

√

1 +

L^2

4 R^2

)

This can readily be differentiated to find the field:

ER=−

dV

dR

= (− 2 kλ)

−L/ 2 R^2 + (1/2)(1 +L^2 / 4 R^2 )−^1 /^2 (−L^2 / 2 R^3 )

L/ 2 R+ (1 +L^2 / 4 R^2 )^1 /^2

,

or, after some simplification,

ER=

kλL

R^2

√

1 +L^2 / 4 R^2

Section 10.3 Fields by superposition 597