Fields of a Ring and a Disk

A ring is the friendliest charge distribution of all. For a point on its axis, every piece of the ring is the same distance away. That makes the integral from the charged rod almost free, and once we have the ring, a disk is just many rings.

Why it works

Take P on the axis, a distance $z$ from the center. Every piece $dq$ is $r = \sqrt{z^2 + R^2}$ from P. Its field has a part along the axis, $dE\cdot\dfrac{z}{r}$, and a part sideways. The piece directly across the ring has an equal and opposite sideways part, so all the sideways parts cancel. The axial parts all point the same way, and since $z$ and $r$ are the same for every piece, they come straight out of the integral: $$E = \frac{z}{r}\cdot\frac{k}{r^2}\int dq = \frac{kQz}{(z^2 + R^2)^{3/2}}.$$

Key idea

On the axis of a ring: $E = \dfrac{kQz}{(z^2 + R^2)^{3/2}}$. On the axis of a disk with surface density $\sigma$: $$E = \frac{\sigma}{2\varepsilon_0}\left(1 - \frac{z}{\sqrt{z^2 + R^2}}\right).$$ Both point along the axis, away from the object if it is positive.

At the center every piece's arrow has a partner pointing the opposite way: the sizes add to a lot, but the vectors add to zero.

Worked examples

Worked example: A ring

A ring of radius $4.0\ \text{cm}$ carries $5.0\ \text{nC}$. Find the field on its axis $3.0\ \text{cm}$ from the center.

$z^2 + R^2 = 0.030^2 + 0.040^2 = 0.0025\ \text{m}^2$, and $(0.0025)^{3/2} = 1.25 \times 10^{-4}\ \text{m}^3$.
$$E = \frac{(8.99 \times 10^{9})(5.0 \times 10^{-9})(0.030)}{1.25 \times 10^{-4}} = 1.08 \times 10^{4}\ \text{N/C}$$

Answer: $E \approx 1.08 \times 10^{4}\ \text{N/C}$ along the axis

Worked example: Where the ring's field peaks

Along the axis, where is the ring's field strongest?

$E(z) \propto z(z^2 + R^2)^{-3/2}$. Differentiate with the product and chain rules: $$\frac{dE}{dz} \propto (z^2 + R^2)^{-3/2} - 3z^2(z^2 + R^2)^{-5/2}.$$
Set it to zero and multiply by $(z^2 + R^2)^{5/2}$: $z^2 + R^2 - 3z^2 = 0$, so $z = R/\sqrt{2}$.
For the $4.0\ \text{cm}$ ring, the peak is $2.8\ \text{cm}$ from the center. The field is zero at the center, rises to this peak, then falls off.

Answer: $z = R/\sqrt{2}$

Worked example: A disk, built from rings

A disk of radius $10\ \text{cm}$ carries $10\ \text{nC}$ spread evenly. Find the field on its axis $2.0\ \text{cm}$ from the center.

A ring of radius $r$ and width $dr$ has $dq = \sigma\,2\pi r\,dr$ and gives $dE = \dfrac{kz\,\sigma\,2\pi r\,dr}{(z^2 + r^2)^{3/2}}$. All of these point the same way, along the axis.
With $u = z^2 + r^2$, $du = 2r\,dr$: $$E = k\sigma\pi z\int_{z^2}^{z^2 + R^2}u^{-3/2}\,du = 2\pi k\sigma\left(1 - \frac{z}{\sqrt{z^2 + R^2}}\right)$$ and $2\pi k = \dfrac{1}{2\varepsilon_0}$.
$\sigma = Q/(\pi R^2) = 3.18 \times 10^{-7}\ \text{C/m²}$, so $\sigma/(2\varepsilon_0) = 1.80 \times 10^{4}\ \text{N/C}$.
$$E = 1.80 \times 10^{4}\left(1 - \frac{0.020}{\sqrt{0.020^2 + 0.10^2}}\right) = 1.45 \times 10^{4}\ \text{N/C}$$

Answer: $E \approx 1.45 \times 10^{4}\ \text{N/C}$

Two limits tell the story of the disk. Very close ($z \ll R$) the bracket is nearly 1 and $E \approx \sigma/(2\varepsilon_0)$, the field of an endless sheet, the same at every small distance. Far away, a binomial expansion of the bracket gives $E \approx kQ/z^2$: the disk looks like a point.

The ring's axial field against distance: zero at the center, a peak at $R/\sqrt{2}$, then a $1/z^2$ fall. Drag P along the axis.

Common mistake

"The field must be huge at the center of the ring, since the charge is so close." It is zero there. Every piece is close, but each one's push is cancelled by the piece across from it. Check with the formula: $z = 0$ makes $E = 0$. (The potential at the center is not zero; potentials are numbers that do not cancel, as you will see in potentials of charge distributions.)

Fields of a Ring and a Disk

Worked examples

Practice

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Sources and further reading