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(D) The interaction energy between two parallel dipoles of moment $p$ separated by distance $r$ is given by $u = -\frac{kp^2}{r^3}$ for anti-parallel

Vedclass · Accepted Answer

$\frac{18U}{17}$

Vedclass · Answer

(D) The interaction energy between two parallel dipoles of moment $p$ separated by distance $r$ is given by $u = -\frac{kp^2}{r^3}$ for anti-parallel alignment and $u = \frac{kp^2}{r^3}$ for parallel alignment.Let the dipoles be $D_1, D_2, D_3$ from left to right.Initial configuration (all parallel): $U_1 = u(D_1, D_2) + u(D_2, D_3) + u(D_1, D_3)$$U_1 = \frac{kp^2}{a^3} + \frac{kp^2}{a^3} + \frac{kp^2}{(2a)^3} = \frac{2kp^2}{a^3} + \frac{kp^2}{8a^3} = \frac{17kp^2}{8a^3} = U$.Final configuration (one end dipole reversed,say $D_1$): $U_2 = u(D_1, D_2) + u(D_2, D_3) + u(D_1, D_3)$$U_2 = -\frac{kp^2}{a^3} + \frac{kp^2}{a^3} - \frac{kp^2}{(2a)^3} = -\frac{kp^2}{8a^3}$.Work done by electric forces $W = U_{initial} - U_{final} = U_1 - U_2$.$W = \frac{17kp^2}{8a^3} - (-\frac{kp^2}{8a^3}) = \frac{18kp^2}{8a^3}$.Since $U = \frac{17kp^2}{8a^3}$,we have $\frac{kp^2}{a^3} = \frac{8U}{17}$.Substituting this into $W$: $W = \frac{18}{8} 	imes \frac{8U}{17} = \frac{18U}{17}$.

Column-$I$ (Angle between $\vec{p}$ and $\vec{E}$)	Column-$II$ (Potential energy of the dipole)
$a. 180^{\circ}$	$i. -pE$
$b. 120^{\circ}$	$ii. pE$
$c. 90^{\circ}$	$iii. \frac{1}{2} pE$
	$iv. 0$

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 dipoles is gradually reversed. How much work is done by the electric forces?

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