Question

1. Two charges are arranged so that a -2.5μC charge is placed at the origin and a 1.7μC charge is placed 0.50m to the right.  What is the electric field at (0.50m, 1.0m)

Answer 1

$E_1=\frac{kq_1}{r_1^2}=\frac{9\cdot 10^9\cdot (-2.5\cdot 10^{-6})}{1.25}=-18~\frac{kN}{C},$

$E_2=\frac{kq_2}{r_2^2}=\frac{9\cdot 10^9\cdot 1.7\cdot 10^{-6}}{1}=15.3~\frac{kN}{C},$

$E=\sqrt{|E_1|^2+|E_2|^2-2|E_1||E_2|cos(atan0.5)}=8.09~\frac{kN}{C}.$