Question

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

Answer 1

The magnitude of the electric field

$$E=k\frac{q}{r^2}$$

$$E=9*10^9*\frac{6.5*10^{-8}}{0.35^2}=4.8*10^3\:\rm N/C$$

The field is directed radially toward the charge.