c/Example 14.
For large values ofR, the square root approaches one, and we
have simplyER≈ kλL/R^2 = k Q/R^2. In other words, the field
very far away is the same regardless of whether the charge is a
point charge or some other shape like a rod. This is intuitively
appealing, and doing this kind of check also helps to reassure
one that the final result is correct.
The preceding example, although it involved some messy alge-
bra, required only straightforward calculus, and no vector operations
at all, because we only had to integrate a scalar function to find the
potential. The next example is one in which we can integrate either
the field or the potential without too much complication.On-axis field of a ring of charge example 14
.Find the potential and field along the axis of a uniformly charged
ring.
.Integrating the potential is straightforward.V=∫
kdq
r=k∫
dq
√
b^2 +z^2
=
k
√
b^2 +z^2∫
dq=
k Q
√
b^2 +z^2,
whereQis the total charge of the ring. This result could have
been derived without calculus, since the distanceris the same
for every point around the ring, i.e., the integrand is a constant.
It would also be straightforward to find the field by differentiating
this expression with respect toz(homework problem 10).
Instead, let’s see how to find the field by direct integration. By
symmetry, the field at the point of interest can have only a com-
ponent along the axis of symmetry, thezaxis:
Ex= 0
Ey= 0
To find the field in thezdirection, we integrate thezcomponents
contributed to the field by each infinitesimal part of the ring.Ez=∫
dEz=∫
|dE|cosθ,whereθis the angle shown in the figure.Ez=∫
kdq
r^2
cosθ=k∫
dq
b^2 +z^2cosθ598 Chapter 10 Fields