Question

Calculate the magnitude and direction of the electric field 0.85 m from a + 3.25 𝜇C point charge. 

Answer 1

The electric field due point charge

$$E=k\frac{q}{r^2}=9\times 10^9\frac{3.25\times 10^{-6}}{0.85^2}=4.05\times 10^4\:\rm N/C$$

The field is directed radially away of charge.