Irodov 3.9: Where the Ring's Field Peaks – Amit & Deepali Sinha Foundation

Problem Irodov 3.9

A thin wire ring of radius $r$ carries a charge $q$ . Find the magnitude $E$ of the electric field on the ring’s axis as a function of the distance $l$ from its centre, and examine how it behaves when $l \gg r$ . Find the largest value of $E$ on the axis and the distance $l$ at which it occurs, and sketch the graph of $E(l)$ .

The thin ring of radius r carrying charge q, its axis, and a point P on the axis at distance l from the centre. — The thin ring of radius $r$ carrying charge $q$ , its axis, and a point $P$ on the axis at distance $l$ from the centre.

Given

ring radius $r$ , charge $q$
point on the axis at distance $l$

Find

E(l)

; its form for

l \gg r

;

E_{\mathrm{max}}

and where it occurs; the graph of

E(l)

Solution

Step 1The field of one element

Assuming the charge spread evenly round the ring, cut it into elements $dq$ . Every element is the same distance from $P$ ; call it $s$ :

s = \sqrt{r^2 + l^2} \tag{1}

One element dq of the ring, its distance s to the axial point P, its field dE at P and the axial component dE_∥. — One element $dq$ of the ring, its distance $s$ to the axial point $P$ , its field $d\vect{E}$ at $P$ and the axial component $dE_{\parallel}$ .

By Coulomb’s law the element gives at $P$ a field of magnitude $dq/(4\pi\varepsilon_0 s^2)$ , directed along the line from the element to $P$ . That line makes with the axis the angle whose cosine is $l/s$ , so the axial component is

dE_{\parallel} = \frac{1}{4\pi\varepsilon_0}\,\frac{dq}{s^2}\cdot\frac{l}{s} = \frac{l\,dq}{4\pi\varepsilon_0 s^3} \tag{2}

Step 2Summing over the ring

In (2) everything but $dq$ is the same for every element, so the sum over the ring simply replaces $dq$ by $q$ . With $s$ from (1):

E = \frac{l}{4\pi\varepsilon_0 s^3}\oint dq = \frac{q\,l}{4\pi\varepsilon_0 s^3}

E = \frac{q\,l}{4\pi\varepsilon_0\,(r^2 + l^2)^{3/2}}

(3)

For $q > 0$ the field points along the axis away from the ring, on both sides of it.

Step 3Far from the ring

For $l \gg r$ , factor $l^2$ out of the bracket in (3) and expand:

\begin{aligned} (r^2 + l^2)^{3/2} &= l^3\left(1 + \frac{r^2}{l^2}\right)^{3/2}\\ E &= \frac{q}{4\pi\varepsilon_0 l^2}\left(1 + \frac{r^2}{l^2}\right)^{-3/2} = \frac{q}{4\pi\varepsilon_0 l^2}\left(1 - \frac{3r^2}{2l^2} + \dots\right) \end{aligned}

Keeping the leading term,

E \approx \frac{q}{4\pi\varepsilon_0\,l^2} \quad (l \gg r)

(4)

From far away the ring is a point charge $q$ at its centre; the first correction is of relative size $r^2/l^2$ .

Step 4The maximum

Near the centre $E$ grows with $l$ , and far away it falls, so it has a maximum in between. Differentiating (3) with respect to $l$ and collecting over the common factor $(r^2 + l^2)^{-5/2}$ :

\begin{aligned} \frac{dE}{dl} &= \frac{q}{4\pi\varepsilon_0}\left[\frac{1}{(r^2 + l^2)^{3/2}} - \frac{3l^2}{(r^2 + l^2)^{5/2}}\right]\\ &= \frac{q}{4\pi\varepsilon_0}\,\frac{r^2 - 2l^2}{(r^2 + l^2)^{5/2}} \end{aligned}

The derivative vanishes at $l^2 = r^2/2$ . At that distance $r^2 + l^2 = 3r^2/2$ , and (3) gives

E_{\mathrm{max}} = \frac{q\,r/\sqrt{2}}{4\pi\varepsilon_0\,(3r^2/2)^{3/2}} = \frac{q\,r}{\sqrt{2}}\cdot\frac{2\sqrt{2}}{3\sqrt{3}\,r^3}\cdot\frac{1}{4\pi\varepsilon_0} = \frac{2q}{3\sqrt{3}\cdot 4\pi\varepsilon_0 r^2}

\begin{gathered}l_m = \frac{r}{\sqrt{2}}\\ E_{\mathrm{max}} = \frac{q}{6\sqrt{3}\,\pi\varepsilon_0 r^2}\end{gathered}

(5)

Step 5The graph

The curve starts at zero at the centre, where the elements’ pulls cancel, rises linearly at first, $E \approx q\,l/(4\pi\varepsilon_0 r^3)$ for $l \ll r$ , peaks at $l_m$ , and then falls toward the point-charge curve $q/(4\pi\varepsilon_0 l^2)$ .

zero at

l = 0

, peak at

l_m = r/\sqrt{2}

, then falling as

1/l^2

Check

The potential on the axis gives the same field. Every element is at distance $s$ , so $\varphi = q/(4\pi\varepsilon_0\sqrt{r^2 + l^2})$ , and

E = -\frac{d\varphi}{dl} = \frac{q}{4\pi\varepsilon_0}\,\frac{l}{(r^2 + l^2)^{3/2}}

which is (3). At $l = 0$ it vanishes, as symmetry demands at the centre.

Remark

Near the centre the axial field grows in proportion to $l$ , like the force of a spring. A small charge of the opposite sign released on the axis near the centre therefore oscillates harmonically along it, with $\omega^2 = \lvert q_0\rvert q/(4\pi\varepsilon_0 m r^3)$ for charge $q_0$ and mass $m$ . The equilibrium is stable only along the axis: in the plane of the ring the same charge is pulled toward the nearer part of the wire, as Earnshaw’s theorem requires.