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Question

(A) The potential energy of interaction between two dipoles with moments $\vec{p}_1$ and $\vec{p}_2$ separated by distance $r$ is given by $U = \frac{

Vedclass · Accepted Answer

$U_1 > U_2 > U_3$

Vedclass · Answer

(A) The potential energy of interaction between two dipoles with moments $\vec{p}_1$ and $\vec{p}_2$ separated by distance $r$ is given by $U = \frac{1}{4\pi\epsilon_0 r^3} [\vec{p}_1 \cdot \vec{p}_2 - 3(\vec{p}_1 \cdot \hat{r})(\vec{p}_2 \cdot \hat{r})]$. For side-by-side dipoles,$\vec{p}_1 \cdot \hat{r} = 0$ and $\vec{p}_2 \cdot \hat{r} = 0$,so $U = \frac{\vec{p}_1 \cdot \vec{p}_2}{4\pi\epsilon_0 r^3}$.In arrangement $(1)$,all three dipoles are parallel. Let $p$ be the magnitude of each dipole. The interaction energy is $U_1 = U_{12} + U_{23} + U_{13} = \frac{p^2}{4\pi\epsilon_0 r^3} + \frac{p^2}{4\pi\epsilon_0 r^3} + \frac{p^2}{4\pi\epsilon_0 (2r)^3} = \frac{p^2}{4\pi\epsilon_0 r^3} (1 + 1 + 1/8) = 2.125 \frac{p^2}{4\pi\epsilon_0 r^3}$. This is positive and the largest.In arrangement $(2)$,the third dipole is anti-parallel. $U_2 = U_{12} + U_{23} + U_{13} = \frac{p^2}{4\pi\epsilon_0 r^3} + \frac{-p^2}{4\pi\epsilon_0 r^3} + \frac{-p^2}{4\pi\epsilon_0 (2r)^3} = \frac{p^2}{4\pi\epsilon_0 r^3} (1 - 1 - 1/8) = -0.125 \frac{p^2}{4\pi\epsilon_0 r^3}$.In arrangement $(3)$,the middle dipole is anti-parallel. $U_3 = U_{12} + U_{23} + U_{13} = \frac{-p^2}{4\pi\epsilon_0 r^3} + \frac{-p^2}{4\pi\epsilon_0 r^3} + \frac{p^2}{4\pi\epsilon_0 (2r)^3} = \frac{p^2}{4\pi\epsilon_0 r^3} (-1 - 1 + 1/8) = -1.875 \frac{p^2}{4\pi\epsilon_0 r^3}$.Comparing the values,$U_1 > U_2 > U_3$.

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