Question

Three identical dipoles with charges $q$ and $-q$ and separation $a$ between the charges are placed at the corners of an equilateral triangle of side $d$ as shown in fig. Find the interaction energy of the system $(a<<d)$.

• A. $\frac{3q^2{a}^2}{4\pi {\epsilon }_0{d}^3}$
• B. $\frac{3q^2{a}^2}{\pi {\epsilon }_0{d}^3}$
• C. $\frac{q^2{a}^2}{7\pi {\epsilon }_0{d}^3}$
• D. $\frac{q^2{a}^2}{2\pi {\epsilon }_0{d}^3}$

Answer 1

The correct option is A $\frac{3q^2{a}^2}{4\pi {\epsilon }_0{d}^3}$


Electric fields $E_1$ and $E_2$ due to dipole at $\text{B}$ are,
$E_1=\frac{k2p\cos {60}^0}{d^3}$

and $E_2=\frac{kp\sin {60}^0}{d^3}$

Potential energy between $\text{A}$ and $\text{B}$ is
$U=-\overrightarrow{p}.\overrightarrow{E}$

$\Rightarrow U=-\overrightarrow{p}.(\overrightarrow{E_1}+\overrightarrow{E_2})$

$\Rightarrow U=-\overrightarrow{p}.\overrightarrow{E_1}(\because \overrightarrow{p}\perp \overrightarrow{E_2})$

$\Rightarrow U=-p{E}_1\cos {180}^0$

$\Rightarrow U=p{E}_1$

Substituting the value of $E_1$,
$\Rightarrow U=\frac{2k{p}^2\cos {60}^0}{d^3}$

As dipole moment, $p=qa$ and $k=\frac{1}{4\pi {\epsilon }_0}$, substituting these in the above equation,

$\Rightarrow U=\frac{2\times \frac{1}{4\pi {\epsilon }_0}\times (qa{)}^2\times \frac{1}{2}}{d^3}$

$\Rightarrow U=\frac{q^2{a}^2}{4\pi {\epsilon }_0{d}^3}$

So, the net energy of system $3U=\frac{3q^2{a}^2}{4\pi {\epsilon }_0{d}^3}$

Hence, option (a) is the correct answer.