E∝1/D³
E∝1/D
E∝1/D⁴
E∝1/D²
Correct option is D. E∝1/D⁴
Electric field at p = 2E₁cosθ₁ - 2E₁cosθ₂ = 2Kq(d² + D²)⁻¹/² × D(d² + D²)⁻¹/² - 2Kq[(2d)² + D²]⁻¹/² × D[(2d)² + D²]⁻¹/²
= 2KqD[(d² + D²)⁻³/² - (4d² + D²)⁻³/²]
= 2KqD/D³[(1 + d²/D²)⁻³/² - (1 + 4d²/D²)⁻³/²]
Applying binomial approximation ≈ d << D
= 2KqD/D³[1 - 3/2d²/D² - (1 - 3/2 × 4d²/D²)]
= 2Kq/D³[1 - 3/2d²/D² - 1 + 6d²/D²] = 9kq d²/D⁴.