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Three identical small electric dipoles are arranged parallel to each other at equal separation a as shown in the figure. Their total interaction energy is $U$. Now one of the end dipole is gradually reversed, how much work is done by the electric forces.
$\frac{17U}{8}$
$\frac{16U}{17}$
$\frac{16U}{8}$
$\frac{18U}{17}$
Solution
Potential energy of a dipole $\mathrm{U}=-\mathrm{p} \overrightarrow{\mathrm{E}}$
Interaction energy and two adjecent dipoles
$\mathrm{u}_{1}=\frac{\mathrm{kp}^{2}}{\mathrm{a}^{3}} \uparrow \uparrow, \mathrm{u}_{2}=-\frac{\mathrm{kp}^{2}}{\mathrm{a}^{3}} \uparrow \downarrow$
interaction energy of the two end dipoles
$\mathrm{u}_{3}=\frac{\mathrm{kp}^{2}}{8 \mathrm{a}^{3}} \uparrow \uparrow ; \quad \mathrm{u}_{4}=-\frac{\mathrm{kp}^{2}}{8 \mathrm{a}^{3}} \uparrow \downarrow$
Total interaction energy in $\mathrm{I}$ configuration
$U_{1}=U=2 u_{1}+u_{3}=\frac{17 k p^{2}}{8 a^{3}}$ ………$(i)$
Total interaction energy in $\mathrm{II}$ configuration
$\mathrm{U}_{2}=\mathrm{u}_{1}+\mathrm{u}_{2}+\mathrm{u}_{4}=\frac{-\mathrm{kp}^{2}}{8 \mathrm{a}^{3}}$ ……..$(ii)$
Work done electric forces
$=U_{1}-U_{2}=\frac{18}{8} \frac{\mathrm{kp}^{2}}{\mathrm{a}^{3}}=\frac{18}{17} \mathrm{U}$
