MCQ
Given below are two statements: one is labelled as Assertion $A$ and the other is labelled as Reason $R$.

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:

  • A
    Both $A$ and $R$ are true and $R$ is $NOT$ the correct explanation of $A$.
  • $A$ is true but $R$ is false.
  • C
    $A$ is false but $R$ is true.
  • D
    Both $A$ and $R$ are true and $R$ is the correct explanation of $A$.

Answer

Correct option: B.
$A$ is true but $R$ is false.
b
The potential $V$ at any point, at distance $r$ from centre of dipole $=\frac{K P \cos \theta}{r^2}$

At axial point where $\theta=0^{\circ}, V=\frac{K P}{r^2}=\frac{9 \times 10^9 \times 4 \times 10^{-6}}{2^2}=9 \times 10^3 \mathrm{~V}$

At axial point where $\theta=180^{\circ}, V=\frac{-K P}{r^2}=-9 \times 10^3 \mathrm{~V}$

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