Answer to Question #309042 in Electric Circuits for kaye

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?

a.0.876 N

b.0.24 N

c.0.04 N

d.0.78 N

Answer 1

Force

$F=F_1+F_2$

$F=\frac{kq_1q_2}{r^2}+\frac{kq_2q_3}{r^2}$

$F=kq_2[\frac{200\times10^{-9}}{100\times10^{-4}}-\frac{50\times10^{-9}}{25\times10^{-4}}]$

$$F=9\times10^9\times100\times10^{-9}[\frac{200\times10^{-9}}{100\times10^{-4}}-\frac{50\times10^{-9}}{25\times10^{-4}}]=0.04N$$

F=0.04N

Option (c) is correct option