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NEET 2024
Exam
NEET
Year
2024
Subject
Physics
Chapter
Electrostatics

NEET 2024 · Physics · Electrostatics

Given below are two statements: one is labelled as Assertion A and...

Given below are two statements: one is labelled as Assertion A\mathbf{A} and the other is labelled as Reason R\mathbf{R}. Assertion A: The potential (V)(V) at any axial point, at 2 m distance (r)(r) from the centre of the dipole of dipole moment vector P⃗\vec{P} of magnitude, 4×10−6Cm4 \times 10^{-6} \mathrm{C} \mathrm{m}, is ±9×103 V\pm 9 \times 10^3 \mathrm{~V}. (Take 14πϵ0=9×109\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 SI units) Reason R\mathbf{R} : V=±2P4πϵ0r2V= \pm \frac{2 P}{4 \pi \epsilon_0 r^2}, where rr 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:

Options

  1. ABoth A and R are true and R is NOT the correct explanation of A .
  2. BAA is true but RR is false.
  3. CAA is false but RR is true.
  4. DBoth AA and RR are true and RR is the correct explanation of AA.

Correct answer

B. AA is true but RR is false.

Step-by-step solution

The potential VV at any point, at distance rr from centre of dipole =KPcos⁡θr2=\frac{K P \cos \theta}{r^2} At axial point where θ=0∘,V=KPr2=9×109×4×10−622=9×103 V\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 θ=180∘,V=−KPr2=−9×103 V\theta=180^{\circ}, V=\frac{-K P}{r^2}=-9 \times 10^3 \mathrm{~V}
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