Question

Three charged particles are located at the corners of a right angled triangle and have charges
q1 = q3 = 40 nC and q2 = 10 nC. Find the magnitude and direction of the electric field at point
P due to these particles.

Answer 1

Electric field from charges 1 and 3 have equal magnitude but opposite direction. So, the net electric field at P is from charge 2

$$E=9\cdot10^{9}\frac{10\cdot10^{-9}}{(0.1\cos{45})^2}=18000\frac{V}{m}$$

Direction:

$$\theta=45\degree$$