Question

Three point charges arranged as shown. q1= 4.0x10-9C, 2 = 5.0x10-9C, q3= -6.0x10-9C. Find the net force on q1 due to q2 and q3.

q2---------- 0.5m ------q1--------- 0.3m -------q3

Answer 1

Force

$F_1=\frac{k q_1q_2}{r^2}$

$$F_1=\frac{9\times10^9\times5\times10^{-9}\times4\times10^{-9}}{0.5^2}=7.2\times10^{-7}N$$

Force

$F_2=\frac{k q_1q_3}{r^2}$

$$F_2=-\frac{9\times10^9\times6\times10^{-9}\times4\times10^{-9}}{0.3^2}=-24\times10^{-7}N\\|F_2|=24\times10^{-7}N$$

Net force

$F_{net}=F_1+F_2$

$$F_{net}=(7.2+24)\times10^{-7}N=31.2\times10^{-7}N$$