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Question

(C) The potential energy of two dipoles $\overrightarrow{p}_{1}$ and $\overrightarrow{p}_{2}$ separated by distance $r$ along the $x$-axis is given by

Vedclass · Accepted Answer

$\frac{p}{a} \sqrt{\frac{1}{2 \pi \varepsilon_{0} ma}}$

Vedclass · Answer

(C) The potential energy of two dipoles $\overrightarrow{p}_{1}$ and $\overrightarrow{p}_{2}$ separated by distance $r$ along the $x$-axis is given by $U = \frac{1}{4\pi\varepsilon_{0}} \left[ \frac{\overrightarrow{p}_{1} \cdot \overrightarrow{p}_{2} - 3(\overrightarrow{p}_{1} \cdot \hat{r})(\overrightarrow{p}_{2} \cdot \hat{r})}{r^{3}} ight]$.Here,$\overrightarrow{p}_{1} = p\hat{i}$,$\overrightarrow{p}_{2} = -p\hat{i}$,and $\hat{r} = \hat{i}$.So,$U = \frac{1}{4\pi\varepsilon_{0}} \left[ \frac{(p\hat{i}) \cdot (-p\hat{i}) - 3(p\hat{i} \cdot \hat{i})(-p\hat{i} \cdot \hat{i})}{a^{3}} ight] = \frac{1}{4\pi\varepsilon_{0}} \left[ \frac{-p^{2} - 3(p)(-p)}{a^{3}} ight] = \frac{1}{4\pi\varepsilon_{0}} \left[ \frac{2p^{2}}{a^{3}} ight] = \frac{p^{2}}{2\pi\varepsilon_{0}a^{3}}$.By conservation of energy,$KE_{i} + PE_{i} = KE_{f} + PE_{f}$.Initially,$KE_{i} = 0$ and $PE_{i} = \frac{p^{2}}{2\pi\varepsilon_{0}a^{3}}$.Finally,at infinite separation,$PE_{f} = 0$ and $KE_{f} = 2 	imes (\frac{1}{2}mv^{2}) = mv^{2}$.Thus,$0 + \frac{p^{2}}{2\pi\varepsilon_{0}a^{3}} = mv^{2} + 0$.$v^{2} = \frac{p^{2}}{2\pi\varepsilon_{0}ma^{3}}$.$v = \frac{p}{a} \sqrt{\frac{1}{2\pi\varepsilon_{0}ma}}$.

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