
$U_{f}=\frac{1}{2} C \varepsilon_{2}^{2}, \Delta U=\frac{1}{2} C\left(\varepsilon_{2}^{2}-\varepsilon_{1}^{2}\right)$
$Q_{i n}=+C \varepsilon_{1}, Q_{f}=-C \varepsilon_{2}, \Delta Q=\left|Q_{j}-Q_{i}\right|$
$=C\left(\varepsilon_{2}+\varepsilon_{1}\right)$
Work done-by battery $W_{b}=\varepsilon_{2} \Delta Q=C\left(\varepsilon_{2}+\varepsilon_{1}\right) \varepsilon_{2}$
Heat generated $=W_{b}-\Delta U$
$=C \varepsilon_{2}^{2}+C \varepsilon_{1} \varepsilon_{2}-\frac{1}{2} C\left(\varepsilon_{2}^{2}-\varepsilon_{1}^{2}\right)$
$=\frac{1}{2} C\left(\varepsilon_{2}^{2}+\varepsilon_{1}^{2}+2 \varepsilon_{1} \varepsilon_{2}\right)=\frac{1}{2} C\left(\varepsilon_{1}+\varepsilon_{2}\right)^{2}$
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:

