Irodov 3.56: Four Charges on a Square – Amit & Deepali Sinha Foundation

Problem Irodov 3.56

Two point charges, $q$ and $-q$ , are a distance $l$ apart, and each is at a distance $l/2$ from an infinite conducting plane. Find:

the magnitude of the electric force on each charge;
the magnitude of the electric field at the midpoint between the two charges.

The charges q and -q, a distance l apart, each at distance l/2 from the conducting plane. — The charges $q$ and $-q$ , a distance $l$ apart, each at distance $l/2$ from the conducting plane.

Given

charges $q$ and $-q$
separation $l$
each $l/2$ from the plane

Find

(a) the force

F

on each charge · (b) the field

E

at the midpoint

Solution

Step 1Two images make the plane an equipotential

Put the plane at $y = 0$ , $q$ at $(-l/2,\, l/2)$ and $-q$ at $(l/2,\, l/2)$ . Each charge gets its mirror image: $q'_1 = -q$ at $(-l/2,\, -l/2)$ under $q$ , and $q'_2 = +q$ at $(l/2,\, -l/2)$ under $-q$ . A point of the plane is equally far from each charge and its image, so each pair contributes zero potential there.

\varphi(x, 0) = \frac{1}{4\pi\varepsilon_0}\left(\frac{q}{r_1} - \frac{q}{r_1}\right) + \frac{1}{4\pi\varepsilon_0}\left(\frac{-q}{r_2} + \frac{q}{r_2}\right) = 0

Here $r_1$ and $r_2$ are the distances from $(x, 0)$ to $q$ and to $-q$ . Above the plane the field of the four charges, which sit at the corners of a square of side $l$ , is the true field.

Step 2The three forces on

q

Write $F_0$ for the Coulomb force between two charges of magnitude $q$ a distance $l$ apart.

F_0 = \frac{1}{4\pi\varepsilon_0}\frac{q^2}{l^2} \tag{1}

The images q'_1 and q'_2 (dashed) complete the square. On q: the pull F_1 of -q, the pull F_2 of q'_1, the push F_3 of q'_2 (all three dashed, to scale) and their resultant F. — The images $q'_1$ and $q'_2$ (dashed) complete the square. On $q$ : the pull $\vect F_1$ of $-q$ , the pull $\vect F_2$ of $q'_1$ , the push $\vect F_3$ of $q'_2$ (all three dashed, to scale) and their resultant $\vect F$ .

The real charge $-q$ and the image $q'_1$ are each a distance $l$ from $q$ and attract it, along $+x$ and along $-y$ ; the image $q'_2$ is across the diagonal, a distance $\sqrt{2}\,l$ , and repels $q$ along the diagonal away from itself.

F_1 = F_2 = F_0, \qquad F_3 = \frac{1}{4\pi\varepsilon_0}\frac{q^2}{(\sqrt{2}\,l)^2} = \frac{F_0}{2}

The diagonal from $q'_2$ to $q$ points along $(-1, 1)/\sqrt{2}$ , so $\vect F_3$ gives components $-F_3/\sqrt{2}$ along $x$ and $+F_3/\sqrt{2}$ along $y$ .

\begin{aligned} F_x &= F_1 - \frac{F_3}{\sqrt{2}} = F_0\left(1 - \frac{1}{2\sqrt{2}}\right) \\ F_y &= -F_2 + \frac{F_3}{\sqrt{2}} = -F_0\left(1 - \frac{1}{2\sqrt{2}}\right) \end{aligned}

Step 3(a) The resultant

The two components are equal in size, so the magnitude is $\sqrt{2}$ times either one; then (1) goes in.

F = \sqrt{2}\,F_0\left(1 - \frac{1}{2\sqrt{2}}\right) = \left(\sqrt{2} - \frac{1}{2}\right)\frac{q^2}{4\pi\varepsilon_0 l^2}

F = \dfrac{(2\sqrt{2} - 1)\,q^2}{8\pi\varepsilon_0 l^2}

(2)

The force on $q$ points at $45^\circ$ below the line of the charges, toward the plane and toward $-q$ . The force on $-q$ is its mirror image in the perpendicular bisector of the pair, so it has the same magnitude.

Step 4(b) The field at the midpoint

The midpoint $O$ is at $(0,\, l/2)$ , a distance $l/2$ from each real charge. The field of $q$ there points away from $q$ , the field of $-q$ toward $-q$ : both along $+x$ , each of size $q/(4\pi\varepsilon_0 (l/2)^2)$ .

E_{\text{real}} = 2\cdot\frac{1}{4\pi\varepsilon_0}\frac{q}{(l/2)^2} = \frac{8}{4\pi\varepsilon_0}\frac{q}{l^2}

Each image sits $l/2$ to one side of $O$ and $l$ below it, so its distance $d$ from $O$ and the cosine of its line with the $x$ axis are

d = \sqrt{\left(\frac{l}{2}\right)^2 + l^2} = \frac{\sqrt{5}}{2}\,l, \qquad \frac{l/2}{d} = \frac{1}{\sqrt{5}}

The vertical parts of the two image fields cancel, and the horizontal parts both point along $-x$ .

E_{\text{images}} = -2\cdot\frac{1}{4\pi\varepsilon_0}\frac{q}{5l^2/4}\cdot\frac{1}{\sqrt{5}} = -\frac{8}{4\pi\varepsilon_0}\frac{q}{l^2}\cdot\frac{1}{5\sqrt{5}}

The field E at the midpoint O of the pair, from the two charges and their images, parallel to the plane and pointing from q toward -q. — The field $\vect E$ at the midpoint $O$ of the pair, from the two charges and their images, parallel to the plane and pointing from $q$ toward $-q$ .

Adding the two, and simplifying $8/(4\pi) = 2/\pi$ :

E = \dfrac{2q}{\pi\varepsilon_0 l^2}\left(1 - \dfrac{1}{5\sqrt{5}}\right)

(3)

The field points along $+x$ , from $q$ toward $-q$ .

Check

Directions. In (a) the resultant must pull $q$ toward the plane (the nearer image is unlike) and toward $-q$ ; $F_x > 0$ and $F_y < 0$ say both. In (b) a reflection in the vertical line through $O$ swaps $q$ with $-q$ and $q'_1$ with $q'_2$ , so it maps the charge layout onto its own negative; the vertical field at $O$ , which that reflection leaves unchanged, must then equal its own negative and vanish, as the cancellation found.

Remark

The plane screens the pair. Without it the force on each charge would be $F_0$ along the line of the charges and the midpoint field $8q/(4\pi\varepsilon_0 l^2)$ . The images form a second pair, reversed, so the mutual pull along the line falls to

F_x = \left(1 - \frac{1}{2\sqrt{2}}\right)F_0 \approx 0.65\,F_0

and the midpoint field drops by the fraction $1/(5\sqrt{5}) \approx 9\%$ . A dipole lying flat against a conductor is partly cancelled by its reversed image, which is why its far field falls off faster than a free dipole’s.