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JEE Main 2025 Jan 23 Shift 1, Physics Q11: Coulomb's Law And Electric Field

JEE Main2025Jan 23, Shift 1Physics
Q.

A point particle of charge Q is located at P along the axis of an electric dipole 1 at a distance as shown in the figure. The point P is also on the equatorial plane of a second electric dipole 2 at a distance . The dipoles are made of opposite charge separated by a distance . For the charge particle at P not to experience any net force, which of the following correctly describes the situation?

  1. A

  2. B

  3. C

  4. D

Solution

For zero net force on Q at P, the axial field from Dipole 1 must cancel the equatorial field from Dipole 2 at the same point.

Axial field of Dipole 1 (taking P at distance from the dipole centre along its axis): , which simplifies to .

Equatorial field of Dipole 2 at distance : .

Setting and cancelling :

.

Squaring and rearranging leads to a transcendental relation:

$ \dfrac{4r^2}{(r^2-a^2)^4} = \dfrac{1}{(r^2+a^2)^3} $

$ \Rightarrow 4r^2(r^2+a^2)^3 = (r^2-a^2)^4 $

$ \Rightarrow 4r^8 \left( 1+\dfrac{a^2}{r^2} \right)^3 = r^8 \left( 1-\dfrac{a^2}{r^2} \right)^4 $

$ \Rightarrow 4 \left( 1+\dfrac{a^2}{r^2} \right)^3 = \left( 1-\dfrac{a^2}{r^2} \right)^4 $

Numerical solution yields .

Note: From exam point of view, it is better to put the values of a/r from the options and check in which case LHS is more close to RHS.

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