$=\frac{1}{2} \times \frac{(3 \times 5) \times 10^{-12} \times(500-300)^{2}}{(3+5) \times 10^{-6}} $
$=\frac{15 \times 10^{-12} \times 4 \times 10^{4}}{2 \times 8 \times 10^{-6}}=0.0375 \mathrm{\,J} $


Assertion $A$: The potential ( $V$ ) at any axial point, at $2 \mathrm{~m}$ distance ( $r$ ) from the centre of the dipole of dipole moment vector $\vec{P}$ of magnitude, $4 \times 10^{-6} \mathrm{C} \mathrm{m}$, is $\pm 9 \times 10^3 \mathrm{~V}$.
(Take $\frac{1}{4 \pi \in_0}=9 \times 10^9 \mathrm{Sl}$ units)
Reason $R$: $V= \pm \frac{2 P}{4 \pi \in_0 r^2}$, where $r$ is the distance of any axial point, situated at $2 \mathrm{~m}$ from the centre of the dipole.
In the light of the above statements, choose the correct answer from the options given below:
