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Step-by-Step Solution
Step 1: Identify the Initial Configuration
Two point dipoles are placed along the x-axis at a separation $a$. Their dipole moments are
$ \vec{p}_1 = p\,\hat{i} $ and $ \vec{p}_2 = -p\,\hat{i} $. Each dipole has mass $m$.
Initially, they are held at rest.
Step 2: Write the Expression for Dipole–Dipole Potential Energy
For two dipoles $ \vec{p}_1 $ and $ \vec{p}_2 $ separated by a vector $ \vec{r} $, the interaction potential energy is
$ U_{12} = \frac{1}{4\pi \epsilon_0} \Biggl[ \frac{\vec{p}_1 \cdot \vec{p}_2}{r^3}
\;-\; 3\,\frac{(\vec{p}_1 \cdot \vec{r})(\vec{p}_2 \cdot \vec{r})}{r^5} \Biggr]. $
Step 3: Substitute the Given Dipoles and Geometry
Here, $ \vec{r} = a\,\hat{i} $. Thus,
$ \vec{p}_1 \cdot \vec{p}_2 = (p\,\hat{i}) \cdot (-p\,\hat{i}) = -p^2. $
$ \vec{p}_1 \cdot \vec{r} = p\,\hat{i} \cdot a\,\hat{i} = p\,a. $
$ \vec{p}_2 \cdot \vec{r} = -p\,\hat{i} \cdot a\,\hat{i} = -p\,a. $
Plugging these into the dipole–dipole energy formula:
$ U_{12} \;=\; \frac{1}{4\pi \epsilon_0}\Biggl[\frac{-p^2}{a^3} \;-\; 3 \,\frac{(p\,a)(-p\,a)}{a^5} \Biggr]
\;=\; \frac{1}{4\pi \epsilon_0}\Bigl[-\frac{p^2}{a^3} + 3 \,\frac{p^2}{a^3}\Bigr]
\;=\; \frac{1}{4\pi \epsilon_0}\,\frac{2\,p^2}{a^3}.
$
Step 4: Use Energy Conservation to Find Final Speed
As the dipoles are released, they repel/attract in such a way that they eventually move infinitely far apart
where the potential energy is zero. By conservation of energy, the initial potential energy transforms into the
total kinetic energy of both dipoles:
$ U_{\text{initial}} + K_{\text{initial}}
\;=\; U_{\text{final}} + K_{\text{final}}. $
Initially, $ K_{\text{initial}} = 0 $ (at rest), and $ U_{\text{final}} = 0 $ (at large separation). Hence,
$ \frac{2\,p^2}{4\pi \epsilon_0\,a^3} \;=\; K_{\text{final}}. $
Since each dipole has the same mass $m$ and acquires the same speed $v$, the final kinetic energy is
$ K_{\text{final}} \;=\; \frac{1}{2}m\,v^2 \;+\; \frac{1}{2}m\,v^2
\;=\; m\,v^2. $
Therefore,
$ m\,v^2
\;=\; \frac{2\,p^2}{4\pi \epsilon_0\,a^3}. $
So,
$ v^2
\;=\; \frac{2\,p^2}{4\pi \epsilon_0\,m\,a^3}
\;=\; \frac{p^2}{2\,\pi \epsilon_0\,m\,a^3}.
$
Taking the square root:
$ v
\;=\; \frac{p}{a^{3/2}}\,\sqrt{\frac{1}{2\,\pi\,\epsilon_0\,m}}
\;=\; \frac{p}{a}\,\sqrt{\frac{1}{2\,\pi\,\epsilon_0\,m\,a}}.
$
Step 5: Match with the Given Options
This simplifies exactly to the correct answer option:
$ \displaystyle \frac{p}{a}\,\sqrt{\frac{1}{2\,\pi\,\epsilon_0\,m\,a}}. $
