Question

What is the magnitude and direction of the electric field at .35 m from a -6.5 x 10^-8 C point charge?

Answer 1

The electric field due point charge

$$E=k\frac{q}{r^2}\\ =\rm9*10^9*\frac{6.5*10^{-8}}{0.25^2}=9360\: N/C$$

Field directed radially toward the charge.