Four point charges –q, +q, +q and –q are placed on y-axis at y = –2d, y = –d, y = +d and y = +2d, respectively. The magnitude of the electric field E at a point on the x-axis at x = D, with D >> d, will behave as :

Engineering

Physics

Electric Field and Charged Particle in Electric Field

Question

$E \propto \frac{1}{D}$

$E \propto \frac{1}{D^{3}}$

$E \propto \frac{1}{D^{2}}$

$E \propto \frac{1}{D^{4}}$

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

Electric field at p = 2E₁cosθ₁ – 2E₂cosθ₂

_{$= \frac{2 Kq}{(d^{2} + D^{2})} \times \frac{D}{(d^{2} + D^{2})} - \frac{2 Kq}{[{(2 d)}^{2} + D^{2}]} \times \frac{D}{{[{(2 d)}^{2} + D^{2}]}^{1 / 2}}$}

= 2KqD [(d² + D²)^–3/2 –(4d² + D²)^–3/2]

$= \frac{2 KqD}{D^{3}} [{(1 + \frac{d^{2}}{D^{2}})}^{- 3 / 2} - {(1 + \frac{4 d^{2}}{2 D^{2}})}^{- 3 / 2}]$

Applying binomial approximation $∵$ d << D

$= \frac{2 KqD}{D^{3}} [1 - \frac{3}{2} \frac{d^{2}}{D^{2}} - (1 + \frac{3 \times 4 d^{2}}{2 D^{2}})]$

$= \frac{2 KqD}{D^{3}} [\frac{12}{2} \frac{d^{2}}{D^{2}} - \frac{3}{2} \frac{d^{2}}{D^{2}}]$

$= \frac{9 {Kqd}^{2}}{D^{4}}$