Question

Find (a) magnitude of the electric field 0.5 m from a 2.5 -μC point charge and (b) the magnitude and direction of the electrostatic force

 acting on an electron placed at that point.

Answer 1

(a) the magnitude electric field due point charge is given by

$$E=k\frac{q}{r^2}=9*10^9*\frac{2.5*10^{-6}}{0.5^2}=9*10^4\:\rm N/C$$

the field is directed radially toward to the charge.

(b) the magnitude of the electrostatic force acting on an electron placed at that point

$$F=eE=1.6*10^{-19}*9*10^4=1.4*10^{-14}\:\rm N$$

the field is directed radially outward from the charge.