Irodov 3.10: A Quadrupole Hides Behind Zero Charge – Amit & Deepali Sinha Foundation

Problem Irodov 3.10

A thin ring of radius $R$ carries a charge $-q$ spread evenly along it, and a point charge $q$ sits at the center of the ring. Find the magnitude of the electric field at a point on the ring’s axis a distance $x$ from the center, for $x \gg R$ .

The ring of radius R carrying -q, the point charge q at its center, and the point P on the axis at distance x; drawn for q > 0. — The ring of radius $R$ carrying $-q$ , the point charge $q$ at its center, and the point $P$ on the axis at distance $x$ ; drawn for $q > 0$ .

Given

$q$ at the center
ring of radius $R$ , charge $-q$ , uniform
$x \gg R$

Find

the magnitude

E

of the field at

P

Solution

Step 1The axial field of a charged ring

Take a ring of total charge $Q$ and the $x$ axis along its axis, from the center toward $P$ . Every element $dq$ of the ring is the same distance from $P$ :

r = \sqrt{x^2 + R^2} \tag{1}

One element dq of the ring, its distance r to P, its field d E at P and the axial component dE_x. For a negative ring d E points back toward the element. — One element $dq$ of the ring, its distance $r$ to $P$ , its field $d\vect E$ at $P$ and the axial component $dE_x$ . For a negative ring $d\vect E$ points back toward the element.

By Coulomb’s law the element’s field at $P$ has magnitude $dq/(4\pi\varepsilon_0 r^2)$ along the line from the element. Its component along the axis carries the cosine of the angle between that line and the axis, which is $x/r$ :

dE_x = \frac{1}{4\pi\varepsilon_0}\,\frac{dq}{r^2}\cdot\frac{x}{r}

Every element shares the same $r$ and $x$ , so the sum over the ring only adds up the charge. With (1):

E_x = \frac{x}{4\pi\varepsilon_0 r^3}\oint dq = \frac{Qx}{4\pi\varepsilon_0 r^3}

E_x = \frac{1}{4\pi\varepsilon_0}\,\frac{Qx}{(x^2+R^2)^{3/2}}

(2)

Step 2Add the charge at the center

The point charge $q$ gives $q/(4\pi\varepsilon_0x^2)$ along $+x$ , and the ring has $Q = -q$ in (2). The two fields lie along the same axis, so they add as signed components:

E_x = \frac{q}{4\pi\varepsilon_0}\left[\frac{1}{x^2} - \frac{x}{(x^2+R^2)^{3/2}}\right] \tag{3}

The two terms nearly cancel far away: the leading $1/x^2$ parts are equal, and only the difference left over by the ring’s size survives.

Step 3Expand for a distant point

Take $x^2$ out of the bracket in the ring’s term, then expand the binomial to first order in the small quantity $R^2/x^2$ :

\begin{aligned} \frac{x}{(x^2+R^2)^{3/2}} &= \frac{1}{x^2}\left(1 + \frac{R^2}{x^2}\right)^{-3/2}\\ &\approx \frac{1}{x^2}\left(1 - \frac{3R^2}{2x^2}\right) \end{aligned}

Substituting into (3), the $1/x^2$ terms cancel:

E_x \approx \frac{q}{4\pi\varepsilon_0}\left[\frac{1}{x^2} - \frac{1}{x^2} + \frac{3R^2}{2x^4}\right]

E = \frac{3qR^2}{8\pi\varepsilon_0 x^4}

(4)

For $q > 0$ the field points along $+x$ , away from the center: the point charge is slightly nearer to $P$ than the ring is, so it wins by a little.

Check

By the potential instead. On the axis the point charge and the ring give

\begin{aligned} \varphi &= \frac{q}{4\pi\varepsilon_0}\left[\frac{1}{x} - \frac{1}{\sqrt{x^2+R^2}}\right] \approx \frac{q}{4\pi\varepsilon_0}\cdot\frac{R^2}{2x^3}\\ E_x &= -\frac{d\varphi}{dx} \approx \frac{q}{4\pi\varepsilon_0}\cdot\frac{3R^2}{2x^4} \end{aligned}

which is (4). Units: $qR^2/x^4$ is $q/x^2$ , a field, times the pure number $(R/x)^2$ . As $R \to 0$ the ring shrinks onto the point charge and the field vanishes, as it must for two coinciding charges $q$ and $-q$ .

Remark

The system has no net charge and, being symmetric under inversion through its center, no dipole moment either, so the first term that survives is the quadrupole one, and it falls as $1/x^4$ . In general the leading field of a localized charge falls as $1/x^2$ for a net charge, $1/x^3$ for a dipole and $1/x^4$ for a quadrupole: each cancellation costs one more power of $R/x$ .