The point charges $5×10^{-11} C$ and $-5×10^{-11} C$ are placed along x-axis at x = 3 cm and x = -3 cm respectively. What will be the electric field intensity at a point (0, 4) cm?

$2.16 × 10^2 N/C$

Electric field due to one charge is $ E = \frac{kq}{r^2} = \frac{9\times 10^9\times 5\times 10^{-11}}{(5\times 10^{-2})^2}= 1.8\times 10^{-2}N/C$

Electric field due to other charge is also same. the angle between two field vector is $(180^o - 74^o)$

Resultant electric field is $ E_{net} = \sqrt{E^2 + E^2+ 2E^2 cos(180^o - 74^o)} = \sqrt{2E^2(1- 0.27)}= \sqrt{1.45 E^2}=1.2E = 2.16 \times 10^{-2}N/C$

The correct answer is Option (2) → $2.16 × 10^2 N/C$