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Electrostatics question

2024 · Q184
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Electrostatics question

2024 · Q184

NEETPhysicsElectrostaticsMCQ+4 / −1

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 m2 \mathrm{~m}2 m distance (r)(r)(r) from the centre of the dipole of dipole moment vector P⃗\vec{P}P of magnitude, 4×10−6Cm4 \times 10^{-6} \mathrm{C} \mathrm{m}4×10−6Cm, is ±9×103 V\pm 9 \times 10^3 \mathrm{~V}±9×103 V.

(Take 14πϵ0=9×109SI\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \mathrm{SI}4πϵ0​1​=9×109SI units)

Reason R: V=±2P4πϵ0r2V= \pm \frac{2 P}{4 \pi \epsilon_0 r^2}V=±4πϵ0​r22P​, where rrr is the distance of any axial point, situated at 2 m2 \mathrm{~m}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. A
    Both A and R are true and R is the correct explanation of A.
  2. B
    Both A and R are true and R is NOT the correct explanation of A.
  3. C
    A is true but R is false.
  4. D
    A is false but R is true.
View written solutionFree

Correct answer: C

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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