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JEE Main 202523 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
23 January
Shift
Morning Shift
Subject
Physics
Chapter
Electrostatics

JEE Main 2025 · 23 January · Morning Shift · Physics · Electrostatics

A point particle of charge Q is located at P along the axis of an...

A point particle of charge QQ is located at PP along the axis of an electric dipole 1 at a distance rr as shown in the figure. The point P is also on the equatorial plane of a second electric dipole 2 at a distance r . The dipoles are made of opposite charge q separated by a distance 2a2 a. For the charge particle at P not to experience any net force, which of the following correctly describes the situation? A point particle of charge Q is located at P along the axis of an electric dipole 1 at a distance r as shown in the

Options

  1. Aar∼10\frac{a}{r} \sim 10
  2. Bar∼20\frac{a}{r} \sim 20
  3. Car∼0.5\frac{a}{r} \sim 0.5
  4. Dar∼3\frac{a}{r} \sim 3

Correct answer

D. ar∼3\frac{a}{r} \sim 3

Step-by-step solution

A point particle of charge Q is located at P along the axis of an electric dipole 1 at a distance r as shown in the
⇒1(r−a)2−1(r+a)2=2a(a2+r2)3/24ar(r2−a2)2=2a(a2+r2)3/2(r2−a2)2=2r(a2+r2)3/2(1−a2r2)2=2(1+a2r2)3/2(1−x2)2=2(1+x2)3/2(x=ar)(1−x2)2(1+x2)3/2=2\begin{aligned} & \Rightarrow \frac{1}{(r-a)^2}-\frac{1}{(r+a)^2}=\frac{2 a}{\left(a^2+r^2\right)^{3 / 2}} \\ & \frac{4 a r}{\left(r^2-a^2\right)^2}=\frac{2 a}{\left(a^2+r^2\right)^{3 / 2}} \\ & \left(r^2-a^2\right)^2=2 r\left(a^2+r^2\right)^{3 / 2} \\ & \left(1-\frac{a^2}{r^2}\right)^2=2\left(1+\frac{a^2}{r^2}\right)^{3 / 2} \\ & \left(1-x^2\right)^2=2\left(1+x^2\right)^{3 / 2}\left(x=\frac{a}{r}\right) \\ & \frac{\left(1-x^2\right)^2}{\left(1+x^2\right)^{3 / 2}}=2 \end{aligned} Now for X=3\mathrm{X}=3 We get 641010≈2⇒ar≈3\frac{64}{10 \sqrt{10}} \approx 2 \Rightarrow \frac{a}{r} \approx 3 [But for a>ra > r point charge will between the dipole where E⃗≠0\vec{E} \neq 0]
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