2024 Electrostatic Potential and CapacitanceElectric Dipole Potential Medium Question 9 of 76 in Electrostatic Potential and Capacitance

Assertion (A): The potential ($V$) at any axial point, at 2 m distance ($r$) from the centre of the dipole of dipole moment vector $\vec{P}$ of magnitude, $4\times10^{-6}$ C m, is $\pm 9\times 10^{3}$ V. (Take $\frac{1}{4\pi\epsilon_0} = 9\times10^{9}$ SI units)

Reason (R): $V = \pm \frac{2P}{4\pi\epsilon_0 r^2}$, where $r$ is the distance of any axial point, situated at 2 m from the centre of the dipole.
A A is true but R is false.
B A is false but R is true.
C Both A and R are true and R is the correct explanation of A.
D Both A and R are true and R is NOT the correct explanation of A.

The answer is C.

Solution

Potential at an axial point due to a dipole: $V = \pm\frac{1}{4\pi\varepsilon_0}\frac{p}{r^2}$ $V = \pm 9 \times 10^9 \times \frac{4 \times 10^{-6}}{(2)^2} = \pm 9 \times 10^3$ V So the Assertion is true. The Reason gives $V = \pm\frac{2p}{4\pi\varepsilon_0 r^2}$, which is not the correct expression (the factor 2 appears in the axial electric field, not the potential). So the Reason is false. Hence A is true but R is false.

Concept Potential at axial point of an electric dipole

Electrostatic Potential and Capacitance in past papers

76 questions from this chapter have appeared across 27 exam years.