Question

find (a) magnitude of the electric field 0.5m 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

Given:

$r=\rm 0.5\: m$

$q=2.5\:\rm μC=2.5*10^{-6}\: C$

The magnitude of electric field

$$E=k\frac{q}{r^2}=\rm 9*10^9*\frac{2.5*10^{-6}\: C}{(0.5\: m)^2}=90000\: V/m$$

The electrostatic force acting on an electron placed at that point

$$F=eE\\ ={\rm 1.6*10^{-19}\: C*90000\: V/m}\\ =1.44*10^{-14}\:\rm N$$

The force is directed toward point charge.