Irodov 3.127: Six Pairs on a Square – Amit & Deepali Sinha Foundation

Problem Irodov 3.127

Four point charges sit at the corners of a square of side $a$ . Find their interaction energy for three arrangements: (a) all four charges are $+q$ ; (b) the charges alternate in sign around the square, $+q$ , $-q$ , $+q$ , $-q$ , so that like charges sit at opposite corners; (c) the two corners along one side carry $+q$ each and the two corners along the opposite side carry $-q$ each.

The three arrangements of charges at the corners of a square of side a; the dashed sides mark the geometry only. — The three arrangements of charges at the corners of a square of side $a$ ; the dashed sides mark the geometry only.

Given

square of side $a$
charges $\pm q$ at the corners, arranged as in (a), (b), (c)

Find

the interaction energy

W

of each arrangement

Solution

Step 1Interaction energy as a sum over pairs

The interaction energy of point charges is the work needed to bring them together from infinite separation, and it is the sum of the Coulomb energies of all distinct pairs:

W = \frac{1}{4\pi\varepsilon_0}\sum_{i<k}\frac{q_iq_k}{r_{ik}} \tag{1}

Four charges make six pairs. Four of them are joined by a side, at distance $a$ ; the other two are joined by a diagonal, at distance $a\sqrt2$ .

Step 2Splitting the sum into sides and diagonals

Grouping the six terms of (1) by distance, with $S$ the sum of the four products $q_iq_k$ along the sides and $D$ the sum of the two products along the diagonals:

W = \frac{1}{4\pi\varepsilon_0}\left(\frac{S}{a} + \frac{D}{a\sqrt2}\right) = \frac{S + D/\sqrt2}{4\pi\varepsilon_0 a} \tag{2}

Only the signs of the products change from one arrangement to the next; each product is $+q^2$ for like charges and $-q^2$ for unlike ones.

Step 3(a) All four charges equal

Every pair is a like pair, so all six products are $+q^2$ :

S = 4q^2, \qquad D = 2q^2

Substituting into (2):

W_a = \frac{4q^2 + 2q^2/\sqrt2}{4\pi\varepsilon_0 a}

With $2/\sqrt2 = \sqrt2$ :

W_a = \frac{(4 + \sqrt2)\,q^2}{4\pi\varepsilon_0 a}

(3)

Step 4(b) Like charges on the diagonals

Each side joins a $+q$ to a $-q$ , and each diagonal joins two equal charges:

S = -4q^2, \qquad D = 2q^2

Substituting into (2):

W_b = \frac{(\sqrt2 - 4)\,q^2}{4\pi\varepsilon_0 a}

(4)

Step 5(c) Like charges along opposite sides

The top side and the bottom side each join equal charges; the two vertical sides and both diagonals join a $+q$ to a $-q$ :

S = q^2 + q^2 - q^2 - q^2 = 0, \qquad D = -2q^2

Substituting into (2):

W_c = -\frac{\sqrt2\,q^2}{4\pi\varepsilon_0 a}

(5)

The side pairs cancel exactly, and the energy comes entirely from the two unlike diagonal pairs.

Check

Each $W$ is $q^2/(4\pi\varepsilon_0 a)$ times a pure number, an energy. A count that cannot be wrong: the sum of all six products is $\left(\sum q_i\right)^2 - \sum q_i^2$ over two. For (a) that is $(16q^2 - 4q^2)/2 = 6q^2$ , and $S + D = 6q^2$ ; for (b) and (c), both neutral, it is $(0 - 4q^2)/2 = -2q^2$ , and indeed $S + D = -4q^2 + 2q^2$ in (b) and $0 - 2q^2$ in (c).

Remark

Arrangement (b) has the lowest energy of the three, $W_b \approx -2.59\,q^2/(4\pi\varepsilon_0 a)$ against $W_c \approx -1.41\,q^2/(4\pi\varepsilon_0 a)$ : putting every unlike pair at the short distance and every like pair at the long one is what an ionic crystal does, and it is why a rock salt lattice alternates its ions in every direction. The negative sign means work must be supplied to pull either neutral square apart.