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NEET PHYSICSMedium

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 m distance (r) from the centre of the dipole of dipole moment vector P of magnitude, 4×10−64\times10^{-6} C m, is ±9×103\pm 9\times10^3 V. (Take 1/4πε0=9×1091/4\pi\varepsilon_0 = 9\times10^9 SI units) Reason (R): V=±2P4πε0r2V = \pm \frac{2P}{4\pi\varepsilon_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).

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).

Step-by-Step Solution

  1. Check Assertion (A): The electric potential VV at an axial point distance rr from a short dipole is given by V=±14πε0Pr2V = \pm \frac{1}{4\pi\varepsilon_0} \frac{P}{r^2}. Given: P=4×10−6P = 4 \times 10^{-6} Cm, r=2r = 2 m, k=9×109k = 9 \times 10^9 Nm2^2/C2^2. Calculation: V=±(9×109)(4×10−6)(2)2=±36×1034=±9×103V = \pm \frac{(9 \times 10^9)(4 \times 10^{-6})}{(2)^2} = \pm \frac{36 \times 10^3}{4} = \pm 9 \times 10^3 V. Thus, Assertion (A) is True.

  2. Check Reason (R): The reason statement gives the formula V=±2P4πε0r2V = \pm \frac{2P}{4\pi\varepsilon_0 r^2}. The correct formula for potential is V=P4πε0r2V = \frac{P}{4\pi\varepsilon_0 r^2}. The factor of 2 in the numerator corresponds to the formula for the Electric Field (E=2P4πε0r3E = \frac{2P}{4\pi\varepsilon_0 r^3}) or is simply incorrect for potential. Thus, Reason (R) is False.

  3. Conclusion: (A) is True but (R) is False.

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