Question

A 100 nC charge is placed at the origin. A -50 nC charge is placed at x= 5 cm and a 200 nC is placed at x= -10 cm.

a. What is the net electrostatic force acting on 100 nC charge?

Answer 1

Net force

$F_{net}=F_1+F_2$

$F_{net}=\frac{kq_1 a_2}{r^2}+\frac{kq_1 q_3}{r^2}$

$$F_{net}={9\times10^9\times100\times10^{-9}}\times(\frac{200\times10^{-9}}{10^{-2}}+\frac{50\times10^{-9}}{0.05^2})$$

$$F_{net}=900\times(2\times10^{-5}+2\times10^{-5})=0.036N$$