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NEET · Physics · STD 12 - 2. Electric potential and capacitance

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

  1. A Both \(A\) and \(R\) are true and \(R\) is \(NOT\) the correct explanation of \(A\).
  2. B \(A\) is true but \(R\) is false.
  3. C \(A\) is false but \(R\) is true.
  4. D Both \(A\) and \(R\) are true and \(R\) is the correct explanation of \(A\).
Verified Solution

Answer & Solution

Correct Answer

(B) \(A\) is true but \(R\) is false.

Step-by-step Solution

Detailed explanation

The potential \(V\) at any point, at distance \(r\) from centre of dipole \(=\frac{K P \cos \theta}{r^2}\) At axial point where \(\theta=0^{\circ}, V=\frac{K P}{r^2}=\frac{9 \times 10^9 \times 4 \times 10^{-6}}{2^2}=9 \times 10^3 \mathrm{~V}\)
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