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PhysicsNEET 2024Potential and potential energy of charges and dipoles

Electrostatic Potential and Capacitance

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A: The potential VV at any axial point, at 2 m distance rr from the centre of a dipole of dipole moment p⃗\vec{p} of magnitude 4×10−64 \times 10^{-6} C m, is ±9×103\pm 9 \times 10^{3} V. (Take 14πε0=9×109\dfrac{1}{4\pi\varepsilon_0} = 9 \times 10^{9} SI units.)

Reason R: V=14πε0 2pr2V = \dfrac{1}{4\pi\varepsilon_0}\,\dfrac{2p}{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. A.Both A and R are true and R is NOT the correct explanation of A.
  2. B.A is true but R is false.✓
  3. C.A is false but R is true.
  4. D.Both A and R are true and R is the correct explanation of A.

Solution

Axial potential of a dipole is V=14πε0pr2V = \frac{1}{4\pi\varepsilon_0}\frac{p}{r^2}, not 2pr2\frac{2p}{r^2} (the factor 2 belongs to the axial field). Numerically V=9×109×4×10−64=9×103V = 9\times10^9 \times \frac{4\times10^{-6}}{4} = 9\times10^3 V, so A is true but R is false.

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