Irodov 3.7: Two Dipoles Side by Side – Amit & Deepali Sinha Foundation

Problem Irodov 3.7

Four point charges sit at the corners of a square whose diagonals have length $2l$ : a charge $q$ at each end of one side and a charge $-q$ at each end of the opposite side. Find the magnitude of the electric field at a point equidistant from all four corners, a distance $x$ from the centre of the square.

The square with diagonals 2l: the two charges +q share the upper side and the two charges -q the lower side. — The square with diagonals $2l$ : the two charges $+q$ share the upper side and the two charges $-q$ the lower side.

Given

charges $\pm q$ at the corners, like charges on one side
diagonal $2l$
point on the axis at distance $x$ from the centre

Find

the magnitude

E

of the field there

Solution

Step 1The point and its distances

A point equidistant from the four corners lies on the line through the centre of the square perpendicular to its plane. Put the origin at the centre of the square, with the unit vector $\vect{i}$ along the two sides that carry like charges, $\vect{j}$ pointing from the $-q$ side toward the $+q$ side, and $\vect{k}$ perpendicular to the square, so that $P$ is at $x\,\vect{k}$ . With $a$ the half side, the charges $+q$ sit at $\pm a\,\vect{i} + a\,\vect{j}$ and the charges $-q$ at $\pm a\,\vect{i} - a\,\vect{j}$ . Each corner is a distance $l$ from the centre, half a diagonal, so

a = \frac{l}{\sqrt{2}} \tag{1}

and every charge is the same distance from $P$ :

r = \sqrt{l^2 + x^2} \tag{2}

The square seen in perspective, with P on the perpendicular through its centre at height x. The two charges +q give E_+ at P, the two charges -q give E_-; their sum E is parallel to the square, from the +q side toward the -q side. — The square seen in perspective, with $P$ on the perpendicular through its centre at height $x$ . The two charges $+q$ give $\vect{E}_+$ at $P$ , the two charges $-q$ give $\vect{E}_-$ ; their sum $\vect{E}$ is parallel to the square, from the $+q$ side toward the $-q$ side.

Step 2The field of each pair

A charge at the corner $\vect{s}$ gives at $P$ a field along the vector $x\,\vect{k} - \vect{s}$ from the charge to $P$ , of length $r$ from (2). For the two positive charges these vectors are $\mp a\,\vect{i} - a\,\vect{j} + x\,\vect{k}$ , and their $\vect{i}$ parts cancel:

\begin{aligned} \vect{E}_+ &= \frac{q}{4\pi\varepsilon_0 r^3}\Big[(-a\,\vect{i} - a\,\vect{j} + x\,\vect{k}) + (a\,\vect{i} - a\,\vect{j} + x\,\vect{k})\Big]\\ &= \frac{q}{4\pi\varepsilon_0 r^3}\,(-2a\,\vect{j} + 2x\,\vect{k}) \end{aligned}

The negative charges have the vectors $\mp a\,\vect{i} + a\,\vect{j} + x\,\vect{k}$ to $P$ and the opposite sign:

\begin{aligned} \vect{E}_- &= -\frac{q}{4\pi\varepsilon_0 r^3}\Big[(-a\,\vect{i} + a\,\vect{j} + x\,\vect{k}) + (a\,\vect{i} + a\,\vect{j} + x\,\vect{k})\Big]\\ &= \frac{q}{4\pi\varepsilon_0 r^3}\,(-2a\,\vect{j} - 2x\,\vect{k}) \end{aligned}

Step 3The total field

Adding the two pairs, only the $\vect{j}$ part remains:

\vect{E} = \vect{E}_+ + \vect{E}_- = -\frac{4qa}{4\pi\varepsilon_0 r^3}\,\vect{j}

Its magnitude, with $a$ from (1) and $r$ from (2), is

E = \frac{4qa}{4\pi\varepsilon_0 r^3} = \frac{4q\,l/\sqrt{2}}{4\pi\varepsilon_0\,(l^2 + x^2)^{3/2}}

E = \frac{q\,l}{\sqrt{2}\,\pi\varepsilon_0\,(l^2 + x^2)^{3/2}}

(3)

The field is parallel to the plane of the square and to the sides joining $+q$ to $-q$ , and points from the side of the positive charges toward the side of the negative ones.

Check

At the centre, $x = 0$ , (3) gives $q/(\sqrt{2}\pi\varepsilon_0 l^2)$ . Directly: each charge is at distance $l$ and gives $q/(4\pi\varepsilon_0 l^2)$ along a diagonal, away from a $+q$ or toward a $-q$ , so all four fields have a component $\cos 45^\circ$ of their length along $-\vect{j}$ and the $\vect{i}$ components cancel in pairs:

E = 4\cdot\frac{1}{\sqrt{2}}\cdot\frac{q}{4\pi\varepsilon_0 l^2} = \frac{q}{\sqrt{2}\,\pi\varepsilon_0 l^2}

the same.

Remark

The square is two parallel dipoles side by side, the pairs $+q$ , $-q$ on the sides joining unlike charges, each of moment $p = q\cdot 2a = \sqrt{2}\,ql$ along $\vect{j}$ , from $-q$ to $+q$ . The axis of the square lies in the equatorial plane of both, where a dipole’s field at distance $x \gg l$ is $p/(4\pi\varepsilon_0 x^3)$ , antiparallel to $\vect{p}$ , so the far field of the pair is

E \approx \frac{2p}{4\pi\varepsilon_0 x^3} = \frac{q\,l}{\sqrt{2}\,\pi\varepsilon_0 x^3}

exactly the limit of (3) for $x \gg l$ . The fall as $1/x^3$ , faster than any single charge’s, is the signature of zero total charge.