Irodov 3.62: A Ring Above Its Own Reflection – Amit & Deepali Sinha Foundation

Problem Irodov 3.62

A thin wire ring of radius $R$ carries a charge $q$ . It lies parallel to an infinite conducting plane, at a distance $l$ from it. Find:

the surface charge density at the point of the plane that lies on the ring’s axis;
the electric field strength and the potential at the center of the ring.

The ring of radius R and charge q, parallel to the conducting plane (seen nearly edge on) at distance l; its axis meets the plane at O. — The ring of radius $R$ and charge $q$ , parallel to the conducting plane (seen nearly edge on) at distance $l$ ; its axis meets the plane at $O$ .

Given

ring radius $R$
charge $q$
parallel to the plane at distance $l$

Find

(a)

\sigma

O

· (b)

E

and

\varphi

at the ring’s center

Solution

Step 1The image ring

The plane is an equipotential at the potential of infinity, zero. Reflect every element $dq$ of the ring in the plane and change its sign: the result is a coaxial ring $-q$ of radius $R$ at depth $l$ , and every point of the plane is equally far from an element and its reflection, so the plane stays at zero. Above the plane the field is that of the two rings.

The field of a ring on its axis is the standard one: by symmetry the components across the axis cancel, and every element is at the same distance $\sqrt{R^2 + z^2}$ from an axial point at distance $z$ , at the same angle.

E(z) = \frac{1}{4\pi\varepsilon_0}\frac{qz}{(R^2 + z^2)^{3/2}} \tag{1}

For $q > 0$ it points along the axis away from the ring.

Step 2(a) The density at

O

The point $O$ is a distance $l$ from each ring on their common axis. The real ring’s field there points away from it, into the metal; the image’s field points toward the image, also into the metal. By (1) the two are equal.

E(O) = 2\cdot\frac{1}{4\pi\varepsilon_0}\frac{ql}{(R^2 + l^2)^{3/2}}

At a conductor’s surface $\sigma = \varepsilon_0 E_n$ , with $E_n$ the component along the outward normal, into the vacuum; here $E_n = -E(O)$ .

\sigma = -\varepsilon_0\cdot\frac{2ql}{4\pi\varepsilon_0 (R^2 + l^2)^{3/2}}

\sigma(O) = -\dfrac{ql}{2\pi (l^2 + R^2)^{3/2}}

(2)

The induced charge at $O$ has the sign opposite to $q$ .

Step 3(b) The field at the center

At its own center the real ring gives no field: its elements’ fields cancel in pairs across the diameter. The field there is the image’s alone, at the axial distance $2l$ , and it points toward the image ring, that is toward the plane.

The image ring -q (dashed), coaxial with the real ring at the mirror position inside the metal. The field E at the center of the real ring is the image ring's field, pointing toward the plane. — The image ring $-q$ (dashed), coaxial with the real ring at the mirror position inside the metal. The field $\vect E$ at the center of the real ring is the image ring’s field, pointing toward the plane.

With $z = 2l$ in (1):

E = \frac{1}{4\pi\varepsilon_0}\frac{q\cdot 2l}{(R^2 + 4l^2)^{3/2}}

E = \dfrac{ql}{2\pi\varepsilon_0 (R^2 + 4l^2)^{3/2}}

(3)

Taking $2l$ out of the bracket writes the same result as

E = \frac{1}{4\pi\varepsilon_0}\frac{q}{4l^2\left[1 + \frac{1}{4}(R/l)^2\right]^{3/2}}

Step 4(b) The potential at the center

The potential is counted from the plane, which is at the potential of infinity. Every element of the real ring is a distance $R$ from the center, and every element of the image a distance $\sqrt{R^2 + 4l^2}$ , so each ring’s potential there is its charge over that one distance.

\varphi = \frac{1}{4\pi\varepsilon_0}\frac{q}{R} + \frac{1}{4\pi\varepsilon_0}\frac{-q}{\sqrt{R^2 + 4l^2}}

\varphi = \dfrac{q}{4\pi\varepsilon_0}\left(\dfrac{1}{R} - \dfrac{1}{\sqrt{R^2 + 4l^2}}\right)

(4)

Check

Shrink the ring to a point, $R \to 0$ . Then (2) becomes $-q/(2\pi l^2)$ , the density under a point charge $q$ at height $l$ (each of the charge and its image gives $q/(4\pi\varepsilon_0 l^2)$ at $O$ ), and (3) becomes the Coulomb field of the image $-q$ at distance $2l$ :

E \to \frac{ql}{2\pi\varepsilon_0 (4l^2)^{3/2}} = \frac{q}{16\pi\varepsilon_0 l^2} = \frac{1}{4\pi\varepsilon_0}\frac{q}{(2l)^2}

And far from the plane, $l \to \infty$ , (4) tends to $q/(4\pi\varepsilon_0 R)$ , the potential at the center of a lone ring.

Remark

The two limits of the potential tell how the plane acts. Far from it, $l \gg R$ , the image is effectively a point charge $-q$ at distance $2l$ and lowers the potential by its Coulomb value; close to it, $l \ll R$ , the ring and its image form a thin double layer and the potential at the center falls to second order in the gap.

\begin{aligned} \varphi &\approx \frac{q}{4\pi\varepsilon_0}\left(\frac{1}{R} - \frac{1}{2l}\right) && (l \gg R) \\ \varphi &\approx \frac{q}{4\pi\varepsilon_0}\,\frac{2l^2}{R^3} && (l \ll R) \end{aligned}