Two identical electric point dipoles have dipole moments $$\vec{p}_1 = p\hat{i}$$ and $$\vec{p}_2 = -p\hat{i}$$ and are held on the x-axis at distance 'a' from each other. When released, they move along the x-axis with the direction of their dipole moments remaining unchanged. If the mass of each dipole is 'm', their speed when they are infinitely far apart is:

First, recall the expression for the electrostatic potential energy of two point dipoles. For dipoles having moments $$\vec p_1$$ and $$\vec p_2$$ separated by the displacement vector $$\vec r$$ (with magnitude $$r$$ and unit vector $$\hat r$$) the formula is

$$U \;=\; \frac1{4\pi\varepsilon_0\,r^{3}}\Bigl[\;\vec p_1\!\cdot\!\vec p_2 \;-\; 3\bigl(\vec p_1\!\cdot\!\hat r\bigr)\bigl(\vec p_2\!\cdot\!\hat r\bigr)\Bigr].$$

We have the two identical dipoles placed on the x-axis a distance $$a$$ apart. Their moments are given as

$$\vec p_1 = +p\hat i, \qquad \vec p_2 = -p\hat i.$$

We choose the displacement vector $$\vec r$$ from dipole 1 to dipole 2 to be along the +x direction, so

$$\vec r = a\hat i, \qquad r = a, \qquad \hat r = \hat i.$$

Now evaluate the scalar products appearing in the energy formula.

1. The direct dot product of the two moments is

$$\vec p_1\!\cdot\!\vec p_2 \;=\; (+p\hat i)\!\cdot\!(-p\hat i) \;=\; -p^{2}.$$

2. The projection of each dipole on the line joining them is

$$\vec p_1\!\cdot\!\hat r \;=\; (+p\hat i)\!\cdot\!\hat i \;=\; +p,$$

$$\vec p_2\!\cdot\!\hat r \;=\; (-p\hat i)\!\cdot\!\hat i \;=\; -p.$$

Multiplying these projections gives

$$\bigl(\vec p_1\!\cdot\!\hat r\bigr)\bigl(\vec p_2\!\cdot\!\hat r\bigr) \;=\; (+p)\times(-p) \;=\; -p^{2}.$$

Substituting all these results into the energy formula:

$$\begin{aligned} U_i &= \frac1{4\pi\varepsilon_0\,a^{3}}\Bigl[\,-p^{2}\;-\;3(-p^{2})\Bigr] \\ &= \frac1{4\pi\varepsilon_0\,a^{3}}\Bigl[-p^{2} + 3p^{2}\Bigr] \\ &= \frac1{4\pi\varepsilon_0\,a^{3}}\,(2p^{2}) \\ &= \frac{2p^{2}}{4\pi\varepsilon_0\,a^{3}} \\ &= \frac{p^{2}}{2\pi\varepsilon_0\,a^{3}}. \end{aligned}$$

This $$U_i$$ is the initial potential energy while both dipoles are at rest, so the initial kinetic energy is zero.

When the dipoles are released they move apart along the x-axis, keeping their orientations fixed. At infinite separation, the dipole-dipole interaction vanishes, so the final potential energy is

$$U_f = 0.$$

By conservation of mechanical energy, the loss in potential energy equals the gain in kinetic energy. Let $$v$$ be the speed acquired by each dipole when they are far apart. Each dipole then has kinetic energy $$\tfrac12 m v^{2}$$, so the total kinetic energy of the pair is

$$K_f = 2\left(\tfrac12 m v^{2}\right) = m v^{2}.$$

Energy conservation gives

$$U_i + 0 = 0 + K_f \quad\Longrightarrow\quad U_i = m v^{2}.$$

Substituting the value of $$U_i$$ calculated earlier, we get

$$\frac{p^{2}}{2\pi\varepsilon_0\,a^{3}} = m v^{2}.$$

Solving for $$v$$:

$$\begin{aligned} v^{2} &= \frac{p^{2}}{2\pi\varepsilon_0\,a^{3} \, m},\\[4pt] v &= \frac{p}{a}\sqrt{\frac{1}{2\pi\varepsilon_0\,m\,a}}. \end{aligned}$$

Thus the speed of each dipole when they are infinitely far apart is

$$v = \frac{p}{a}\sqrt{\frac{1}{2\pi\varepsilon_0\,m\,a}}.$$

Hence, the correct answer is Option B.