23.4.1 A Disk of Charge

FIGURE 23.16 shows a disk of radius R that is uniformly charged with charge Q. This is a mathematical disk, with no thickness, and its surface charge density is

η = \frac{Q}{A} = \frac{Q}{π R^{2}}

(23.18)

We would like to calculate the on-axis electric field at distance z from the center of this disk. Our problem-solving strategy tells us to divide a continuous charge into segments for which we already know how to find $\vec{E}$ . Because we now know the on-axis electric field of a ring of charge, let’s divide the disk into N very narrow rings of radius r and width $Δ r$ . One such ring, with radius $r_{i}$ and charge $Δ Q_{i}$ , is shown.

The figure illustrates the on-axis field of a charged disk.

FIGURE 23.16 Full Alternative Text

We need to be careful with notation. The R in Example 23.4 was the radius of the ring. Now we have many rings, and the radius of ring i is $r_{i}$ . Similarly, Q was the charge on the ring. Now the charge on ring i is $Δ Q_{i}$ , a small fraction of the total charge on the disk. With these changes, the electric field at distance z of ring i, with radius $r_{i}$ , is

{(E_{i})}_{z} = \frac{1}{4 π ϵ_{0}} \frac{z Δ Q_{i}}{{(z^{2} + r_{i}^{2})}^{3 / 2}}

(23.19)

The on-axis electric field of the charged disk is the sum of the electric fields of all of the rings:

{(E_{disk})}_{z} = \sum_{i = 1}^{N} {(E_{i})}_{z} = \frac{z}{4 π ϵ_{0}} \sum_{i = 1}^{N} \frac{Δ Q_{i}}{{(z^{2} + r_{i}^{2})}^{3 / 2}}

(23.20)

The critical step, as always, is to relate $Δ Q$ to a coordinate. Because we now have a surface, rather than a line, the charge in ring i is $Δ Q = η Δ A_{i}$ , where $Δ A_{i}$ is the area of ring i. We can find $Δ A_{i}$ , as you’ve learned to do in calculus, by “unrolling” the ring to form a narrow rectangle of length $2 π r_{i}$ and width $Δ r$ . Thus the area of ring i is $Δ A_{i} = 2 π r_{i} Δ r$ and the charge is $Δ Q_{i} = 2 π η r_{i} Δ r$ . With this substitution, Equation 23.20 becomes

{(E_{disk})}_{z} = \frac{η z}{2 ϵ_{0}} {\sum^{​}}_{i = 1}^{N} \frac{r_{i} Δ r}{{(z^{2} + r_{i}^{2})}^{3 / 2}}

(23.21)

As $N \to \infty$ , $Δ r \to d r$ and the sum becomes an integral. Adding all the rings means integrating from $r = 0$ to $r = R;$ thus

(E_{disk})_{z} = \frac{η z}{2 ϵ_{0}} \int_{0}^{R} \frac{r d r}{{(z^{2} + r^{2})}^{3 / 2}}

(23.22)

All that remains is to carry out the integration. This is straightforward if we make the variable change $u = z^{2} + r^{2}$ . Then $d u = 2 r d r$ or, equivalently, $r d r = \frac{1}{2} d u$ . At the lower integration limit $r = 0$ , our new variable is $u = z^{2}$ . At the upper limit $r = R$ , the new variable is $u = z^{2} + R^{2}$ .

NOTE When changing variables in a definite integral, you must also change the limits of integration.

With this variable change the integral becomes

{(E_{disk})}_{z} = \frac{η z}{2 ϵ_{0}} \frac{1}{2} {\int_{z^{2}}^{z^{2} + R^{2}} \frac{d u}{u^{3 / 2}} = \frac{η z}{4 ϵ_{0}} \frac{- 2}{u^{1 / 2}} |}_{z^{2}}^{z^{2} + R^{2}} = \frac{η z}{2 ϵ_{0}} [\frac{1}{z} - \frac{1}{\sqrt{z^{2} + R^{2}}}]

(23.23)

If we multiply through by z, the on-axis electric field of a charged disk with surface charge density $η = Q / π R^{2}$ is

(E_{disk})_{z} = \frac{η}{2 ϵ_{0}} [1 - \frac{z}{\sqrt{z^{2} + R^{2}}}]

(23.24)

NOTE This expression is valid only for $z > 0$ . The field for $z < 0$ has the same magnitude but points in the opposite direction.