Question

Calculate the magnitude and direction of the electric field at a point P which is 30 cm to the right of a point charge Q = -3.0 x 10^-6 C

Answer 1

The electric field due the point charge

$$E=k\frac{q}{r^2}=9*10^9*\frac{3.0*10^{-6}}{(0.3)^2}=3*10^5\:\rm V/m$$

Direction is toward the charge.