Physics · Electrostatics

NEET (UG) 2024 — Question 5

Given below are two statements: one is labelled as Assertion 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: 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:

  1. Option A:

    Both AA and RR are true and RR is the correct explanation of AA.

  2. Option B:

    Both A and R are true and R is NOT the correct explanation of A .

  3. Option C:

    AA is true but RR is false.

    Correct
  4. Option D:

    AA is false but RR is true.

Answer: C

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}

Answer key and solution verified before publishing.

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Exam
NEET (UG) 2024
Subject
Physics
Chapter
Electrostatics
Topic
Electric Dipole
Given below are two statements: one is labelled as Assertion A and… | NEET (UG) 2024 PYQ with Solution · DhiX AI