Q105 / 200 · NEET 2024 · Physics

Given below are two statements: one is labelled as Assertion $\mathbf{A}$ and the other is labelled as Reason $\mathbf{R}$. Assertion A: The potential $(V)$ at any axial point, at 2 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 $\frac{1}{4 \pi \epsilon_0}=9 \times 10^9$ SI units) Reason $\mathbf{R}$ : $V= \pm \frac{2 P}{4 \pi \epsilon_0 r^2}$, where $r$ is the distance of any axial point, situated at 2 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 .
  • Correct answer
    B
    $A$ 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 $A$.

Solution

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

Hence, the correct answer is option: 'B'.

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