Question

Two positive point charges, each 16 μC , lie along the x-axis at x = –0.15 m and x = +0.15 m . Find the electric field at the point (0, 0.22 m ) on the y-axis.

Answer 1

r₁ = distance from charge at x = 0.15 = sqrt(0.15² + 0.22²) = 0.266 m

r₂ = distance from charge at x = - 0.15 = sqrt(0.15² + 0.22²) = 0.266 m

$\theta$ = tan^-1(0.22/0.15) = 55.71

q = charge = 16 x 10^-6 C

Electric field components along X-direction cancel out and net electric field along Y-direction is given as

E_net = k q Sin55.71/r₁² + k qSin55.71/r₂² = 2 (9 x 10⁹) (16 x 10^-6) Sin55.71/(0.266)² = 3.4 x 10⁶ N/C