EXAMPLE 21.11 FIELD OF A UNIFORMLY CHARGED DISK
A nonconducting disk of radius R has a uniform positive surface
charge density o. Find the electric field at a point along the axis of
the disk a distance x from its center. Assume that x is positive.
SOLUTION
IDENTIFY and SET UP: Figure 21.25 shows the situation. We rep-
resent the charge distribution as a collection of concentric rings of
charge dQ. In Example 21.9 we obtained Eq. (21.8) for the field on
the axis of a single uniformly charged ring, so all we need do here
is integrate the contributions of our rings.
EXECUTE: A typical ring has charge dQ, inner radius r, and outer
radius r + dr. Its area is approximately equal to its width dr times
its circumference 2πr, or dA = 2πr dr. The charge per unit area is
σ = dQ/dA, so the charge of the ring is do = o da = 2πor dr.
We use dQ in place of Q in Eq. (21.8), the expression for the field
due to a ring that we found in Example 21.9, and replace the ring
21.25 Our sketch for this problem.
Idr
dQ
X
P
dEx
dEx
radius a with r. Then the field component dEx at point P due to
this ring is
1
2πσrx dr
4π€0 (x² + ²)3/2
SOLUTION
Continued
