Irodov 3.12: A Cosine Ring Is a Dipole – Amit & Deepali Sinha Foundation

Problem Irodov 3.12

A thin insulating ring of radius $R$ carries charge with linear density $\lambda = \lambda_0 \cos\varphi$ , where $\lambda_0$ is a constant and $\varphi$ is the azimuthal angle around the ring. Find the magnitude of the electric field:

at the center of the ring;
on the ring’s axis, as a function of the distance $x$ from the center, and examine how it behaves for $x \gg R$ .

The ring of radius R with λ = λ_0φ, positive on the side where φ = 0 and negative opposite, its center O, and the point P on the axis at distance x; drawn for λ_0 > 0. — The ring of radius $R$ with $\lambda = \lambda_0\cos\varphi$ , positive on the side where $\varphi = 0$ and negative opposite, its center $O$ , and the point $P$ on the axis at distance $x$ ; drawn for $\lambda_0 > 0$ .

Given

ring of radius $R$
$\lambda = \lambda_0\cos\varphi$

Find

the magnitude of

\vect E

(a) at the center, (b) on the axis at distance

x

, and its form for

x \gg R

Solution

Step 1The field of one element

Put the ring in the $yz$ plane with its center at the origin and the $x$ axis along its axis, and read $\varphi$ from the $y$ axis toward $z$ . The element at angle $\varphi$ sits at $R(0, \cos\varphi, \sin\varphi)$ , spans the length $R\,d\varphi$ and carries

dq = \lambda R\,d\varphi = \lambda_0 R\cos\varphi\,d\varphi \tag{1}

One element dq of the ring, its distance r to the axial point P, and its field d E there. — One element $dq$ of the ring, its distance $r$ to the axial point $P$ , and its field $d\vect E$ there.

Seen from the axial point $P = (x, 0, 0)$ , every element is at the same distance $r = \sqrt{x^2+R^2}$ , and the vector from the element to $P$ is $(x, -R\cos\varphi, -R\sin\varphi)$ . Coulomb’s law gives the element’s field as $dq$ times that vector over $4\pi\varepsilon_0 r^3$ ; with (1) its components are

\begin{aligned} dE_x &= \frac{\lambda_0 R x}{4\pi\varepsilon_0 r^3}\cos\varphi\,d\varphi\\ dE_y &= -\frac{\lambda_0 R^2}{4\pi\varepsilon_0 r^3}\cos^2\varphi\,d\varphi\\ dE_z &= -\frac{\lambda_0 R^2}{4\pi\varepsilon_0 r^3}\sin\varphi\cos\varphi\,d\varphi \end{aligned}

Step 2Add the elements round the ring

Integrating over one turn, $\varphi$ from $0$ to $2\pi$ , the first and last components vanish:

\begin{aligned} \int_0^{2\pi}\cos\varphi\,d\varphi &= \Big[\sin\varphi\Big]_0^{2\pi} = 0\\ \int_0^{2\pi}\sin\varphi\cos\varphi\,d\varphi &= \left[\frac{\sin^2\varphi}{2}\right]_0^{2\pi} = 0 \end{aligned}

The $y$ component survives, using $\cos^2\varphi = (1 + \cos 2\varphi)/2$ :

\int_0^{2\pi}\cos^2\varphi\,d\varphi = \left[\frac{\varphi}{2} + \frac{\sin 2\varphi}{4}\right]_0^{2\pi} = \pi

So the field on the axis points along $-y$ , from the positive half of the ring toward the negative half, with

E_y = -\frac{\lambda_0 R^2}{4\pi\varepsilon_0 r^3}\cdot\pi = -\frac{\lambda_0 R^2}{4\varepsilon_0 (x^2+R^2)^{3/2}} \tag{2}

Step 3(a) The center of the ring

Setting $x = 0$ in (2), the denominator becomes $R^3$ :

E_y = -\frac{\lambda_0 R^2}{4\varepsilon_0 R^3}

E = \frac{\lambda_0}{4\varepsilon_0 R}

(3)

Step 4(b) Along the axis

The magnitude of (2) at any distance $x$ is

E = \frac{\lambda_0 R^2}{4\varepsilon_0 (x^2 + R^2)^{3/2}}

(4)

It decreases monotonically from the value at the center. For $x \gg R$ the $R^2$ beside $x^2$ can be dropped:

E \approx \frac{\lambda_0 R^2}{4\varepsilon_0 x^3} = \frac{\pi R^2\lambda_0}{4\pi\varepsilon_0 x^3}

The numerator is the ring’s dipole moment, $\vect p = \oint \vect r\,dq$ . Its $y$ component, with (1) and the integral of $\cos^2\varphi$ above, is

p = \int_0^{2\pi} R\cos\varphi\cdot\lambda_0R\cos\varphi\,d\varphi = \pi R^2\lambda_0

while its $x$ and $z$ components vanish by the same two integrals as before. Far away the ring is a point dipole:

E \approx \frac{p}{4\pi\varepsilon_0 x^3},\quad p = \pi R^2\lambda_0

(5)

Check

Units: $\lambda_0/(\varepsilon_0 R)$ is $(\lambda_0 R)/(\varepsilon_0R^2)$ , a charge divided by $\varepsilon_0$ and by an area, which is a field. The far form, by a second route: treat each half of the ring as its charge placed at its charge centroid. The half with $\cos\varphi > 0$ carries $Q$ and has its centroid at distance $d$ from $O$ along $+y$ , the other half the opposite:

\begin{aligned} Q &= \int_{-\pi/2}^{\pi/2}\lambda_0R\cos\varphi\,d\varphi = 2\lambda_0R\\ d &= \frac{1}{Q}\int_{-\pi/2}^{\pi/2}R\cos\varphi\cdot\lambda_0R\cos\varphi\,d\varphi = \frac{\pi R}{4} \end{aligned}

so the pair has moment $Q\cdot 2d = \pi R^2\lambda_0$ , the $p$ of (5). A point dipole’s field on the line through it perpendicular to $\vect p$ is $-\vect p/(4\pi\varepsilon_0 x^3)$ , which matches (5) in size and (2) in direction: along $-y$ , against $\vect p$ .

Remark

The ring has no net charge, so its field cannot fall as $1/x^2$ ; the $\cos\varphi$ distribution is pure dipole, and the exact axial field $\lambda_0R^2/[4\varepsilon_0(x^2+R^2)^{3/2}]$ is just the far dipole form with $x^3$ replaced by $(x^2+R^2)^{3/2}$ , the cube of the true distance to every element. So the field is finite even at the center, and at every point of the axis it is perpendicular to the axis.