STD 12 - 2. Electric potential and capacitance — NEET STD 11 Science — Question

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 $\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \mathrm{Sl}$ units)

Reason $R$: $V= \pm \frac{2 P}{4 \pi \epsilon_0 r^2}$, 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: