23.4.2 Limiting Cases

It’s a bit difficult to see what Equation 23.24 is telling us, so let’s compare it to what we already know. First, you can see that the quantity in square brackets is dimensionless. The surface charge density $η = Q / A$ has the same units as $q / r^{2}$ , so $η / 2 ϵ_{0}$ has the same units as $q / 4 π ϵ_{0} r^{2}$ . This tells us that $η / 2 ϵ_{0}$ really is an electric field.

Next, let’s move very far away from the disk. At distance $z ≫ R$ , the disk appears to be a point charge Q in the distance and the field of the disk should approach that of a point charge. If we let $z \to \infty$ in Equation 23.24, so that $z^{2} + R^{2} \approx z^{2}$ , we find $(E_{disk})_{z} \to 0$ . This is true, but not quite what we wanted. We need to let z be very large in comparison to R, but not so large as to make $E_{disk}$ vanish. That requires a little more care in taking the limit.

We can cast Equation 23.24 into a somewhat more useful form by factoring the $z^{2}$ out of the square root to give

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

(23.25)

Now $R^{2} / z^{2} ≪ 1$ if $z ≫ R$ , so the second term in the square brackets is of the form ${(1 + x)}^{- 1 / 2}$ where $x ≪ 1$ . We can then use the binomial approximation

{(1 + x)}^{n} \approx 1 + n x if x ≪ 1 (binomial approximation)

to simplify the expression in square brackets:

1 - \frac{1}{\sqrt{1 + R^{2} / z^{2}}} = 1 - {(1 + R^{2} / z^{2})}^{- 1 / 2} \approx 1 - (1 + (- \frac{1}{2}) \frac{R^{2}}{z^{2}}) = \frac{R^{2}}{2 z^{2}}

(23.26)

This is a good approximation when $z ≫ R$ . Substituting this approximation into Equation 23.25, we find that the electric field of the disk for $z ≫ R$ is

(E_{disk})_{z} \approx \frac{η}{2 ϵ_{0}} \frac{R^{2}}{2 z^{2}} = \frac{Q / π R^{2}}{4 ϵ_{0}} \frac{R^{2}}{z^{2}} = \frac{1}{4 π ϵ_{0}} \frac{Q}{z^{2}} if z ≫ R

(23.27)

This is, indeed, the field of a point charge Q, giving us confidence in Equation 23.24 for the on-axis electric field of a disk of charge.

EXAMPLE 23.5 The electric field of a charged disk

A 10-cm-diameter plastic disk is charged uniformly with an extra $10^{11}$ electrons. What is the electric field 1.0 mm above the surface at a point near the center?

MODEL Model the plastic disk as a uniformly charged disk. We are seeking the on-axis electric field. Because the charge is negative, the field will point toward the disk.

SOLVE The total charge on the plastic square is $Q = N (- e) = - 1.60 \times 10^{- 8} C$ . The surface charge density is

η = \frac{Q}{A} = \frac{Q}{π R^{2}} = \frac{- 1.60 \times 10^{- 8} C}{π (0.050 m)^{2}} = - 2.04 \times 10^{- 6} C / m^{2}

The electric field at $z = 0.0010 m,$ given by Equation 23.25, is

E_{z} = \frac{η}{2 ϵ_{0}} [1 - \frac{1}{\sqrt{1 + R^{2} / z^{2}}}] = - 1.1 \times 10^{5} N / C

The minus sign indicates that the field points toward, rather than away from, the disk. As a vector,

\vec{E} = (1.1 \times 10^{5} N / C, t o w a r d t h e d i s k)

REVIEW The total charge, $- 16 nC,$ is typical of the amount of charge produced on a small plastic object by rubbing or friction. Thus $10^{5} N / C$ is a typical electric field strength near an object that has been charged by rubbing, as you saw in Table 23.1.