Answer to Question #310044 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.

b. What is the net electrostatic force acting on 200nC charge?

Answer 1

Solution;

$F=\frac {q_1q_2}{r^2}$

$F_{200}=(\frac{100×200}{10^2}-\frac{100×50}{5^2})×8.99×10^9$

$F_{200}=0.03596N$