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NEET 2024 · Electric Dipole

Electric Dipole question from NEET 2024

Q30NEET 2024Easy

Electric Dipole

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

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

Reason R: V=±2P4πϵ0r2V = \pm\frac{2P}{4\pi\epsilon_0 r^2}, where rr is the distance of any axial point, situated at 2 m2\text{ m} from the centre of the dipole.

In the light of the above statements, choose the correct answer from the options given below:

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Answer: (c)

  1. Axial potential: V=±P4πϵ0r2V = \pm\frac{P}{4\pi\epsilon_0 r^2} (sign depends on which side of the dipole).
  2. V=9×109×4×10−622=9×103V = \frac{9\times10^9\times4\times10^{-6}}{2^2} = 9\times10^3 V, so A is true.
  3. R has an extra factor 2 (that belongs to the axial field), so R is false.
  4. So the answer is (c).
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