JEE Main20199 Apr 2019Evening ShiftPhysicsElectrostaticsActual
Four point charges - q , + q , + q and - q are placed on y-axis at y = - 2 d , y = - d , and y = + 2 d , respectively. The magnitude of the electric field E at a point on the x-axis at x = D , with D ≫ d , will behave as:
Options
- AE ∝ 1 D 4
- BE ∝ 1 D
- CE ∝ 1 D 3
- DE ∝ 1 D 2
Correct answer
A. E ∝ 1 D 4
Step-by-step solution
The electric field at P is superposition of the fields due to all four charges. The electric field due to the two + q charges is E + q = K q D 2 + d 2 cos ⁡ θ 1 × 2 E - q = K q D 2 + 4 d 2 cos ⁡ θ 2 × 2 The field due to + q is towards right and the field due to - q is towards left. Also θ 1 ≈ 0 and θ 2 ≈ 0 . ∴ E n e t = E + q - E - q ≈ K q D 2 + d 2 - K q D 2 + 4 d 2 = K q   3 d 2 D 2 + d 2 D 2 + 4 d 2 ≈ 3 K q d 2 D 4 E ∝ 1 D 4