Irodov 3.57: The Third Image Pushes Back – Amit & Deepali Sinha Foundation

Problem Irodov 3.57

Two conducting half-planes meet along a common edge at a right angle, forming a corner. A point charge $q$ sits inside the right-angled corner, at the same distance $l$ from each half-plane. Find the magnitude of the force on the charge.

The conducting corner (steel) seen along its edge, with the charge q at distance l from each face. — The conducting corner (steel) seen along its edge, with the charge $q$ at distance $l$ from each face.

Given

point charge $q$
$l$ from each face
faces at $90^\circ$

Find

F = |\vect F|

Solution

Step 1Three images make both faces equipotential

Put the edge of the corner along the $z$ axis, with the faces on the half-planes $x = 0$ ( $y > 0$ ) and $y = 0$ ( $x > 0$ ), so that $q$ sits at $(l, l)$ . Reflecting $q$ in the face $x = 0$ gives $-q$ at $(-l, l)$ ; reflecting that pair in the face $y = 0$ adds $-q$ at $(l, -l)$ and $+q$ at $(-l, -l)$ . With $r(a,b)$ the distance from $(x,y)$ to the point $(a,b)$ , the four charges give the potential

\varphi(x,y) = \frac{q}{4\pi\varepsilon_0}\left[\frac{1}{r(l,l)} - \frac{1}{r(-l,l)} - \frac{1}{r(l,-l)} + \frac{1}{r(-l,-l)}\right] \tag{1}

On the face $x = 0$ a point is equally far from $(l, b)$ and $(-l, b)$ , so the terms of (1) cancel in pairs.

r(l,l) = r(-l,l), \quad r(l,-l) = r(-l,-l) \quad\text{at } x = 0

The same pairing in $y$ cancels the terms on the face $y = 0$ , so both faces sit at $\varphi = 0$ .

Step 2The three partial forces

Name the images as in the figure: $q'_1 = -q$ at $(l,-l)$ , $q'_2 = +q$ at $(-l,-l)$ , $q'_3 = -q$ at $(-l,l)$ . Their distances from $q$ are $2l$ , $2\sqrt{2}\,l$ and $2l$ .

The images q'_1, q'_2, q'_3 (dashed) at their true positions inside the metal. The partial forces F_1, F_2, F_3 (dashed) and their resultant F act on q, drawn to scale. — The images $q'_1$ , $q'_2$ , $q'_3$ (dashed) at their true positions inside the metal. The partial forces $\vect F_1$ , $\vect F_2$ , $\vect F_3$ (dashed) and their resultant $\vect F$ act on $q$ , drawn to scale.

Coulomb’s law with each distance gives the three magnitudes.

F_1 = F_3 = \frac{1}{4\pi\varepsilon_0}\frac{q^2}{(2l)^2} = \frac{q^2}{16\pi\varepsilon_0 l^2} \tag{2}

F_2 = \frac{1}{4\pi\varepsilon_0}\frac{q^2}{(2\sqrt{2}\,l)^2} = \frac{q^2}{32\pi\varepsilon_0 l^2} \tag{3}

$\vect F_1$ and $\vect F_3$ are attractions, perpendicular to each other and pointing at the two faces; $\vect F_2$ is a repulsion along the diagonal, directly away from the edge.

Step 3The resultant along the bisector

The two equal attractions are at right angles, so their sum has magnitude $\sqrt{2}\,F_1$ and lies along the bisector toward the edge, exactly opposite $\vect F_2$ . The magnitudes along that line subtract, using (2) and (3).

\begin{aligned} F &= \sqrt{2}\,F_1 - F_2 \\ &= \frac{\sqrt{2}\,q^2}{16\pi\varepsilon_0 l^2} - \frac{q^2}{32\pi\varepsilon_0 l^2} \end{aligned}

F = \dfrac{(2\sqrt{2}-1)\,q^2}{32\pi\varepsilon_0 l^2}

(4)

The force points along the bisector of the corner, toward its edge.

Check

The units are those of $q^2/(\varepsilon_0 l^2)$ , a force. By components instead of along the bisector: only $q'_3$ (attracting, at $2l$ ) and $q'_2$ (repelling, at $2\sqrt{2}\,l$ , its line at $45^\circ$ ) push along $x$ , and $F_y$ equals $F_x$ by the mirror symmetry of the corner.

F_x = -\frac{q^2}{4\pi\varepsilon_0 (2l)^2} + \frac{q^2}{4\pi\varepsilon_0 (2\sqrt{2}\,l)^2}\frac{1}{\sqrt{2}} = -\frac{q^2}{4\pi\varepsilon_0 l^2}\left(\frac{1}{4} - \frac{1}{8\sqrt{2}}\right)

F = \sqrt{2}\,|F_x| = \frac{q^2}{4\pi\varepsilon_0 l^2}\left(\frac{\sqrt{2}}{4} - \frac{1}{8}\right) = \frac{(2\sqrt{2}-1)\,q^2}{32\pi\varepsilon_0 l^2}

which is (4). The sign is right: the two unlike images are nearer than the like one, so the net force pulls $q$ toward the metal.

Remark

The corner pulls less than a single face would. One plane at distance $l$ attracts with $q^2/(16\pi\varepsilon_0 l^2) = 0.0625\,q^2/(\pi\varepsilon_0 l^2)$ , and the corner’s $(2\sqrt{2}-1)/32 = 0.0571$ in the same units is about $9\%$ weaker, because the third image, a like charge, pushes back along the very line the two attractions combine on. The construction closes only for corners of angle $\pi/n$ , which need $2n-1$ images; for any other angle some reflected image lands in the vacuum region itself, where it would be a charge that is not there.