Question

three charges are located at 100m intervals along a horizontal line: a charge of -3.0C on the left, 2.0C in the middle, and 1.0C on the right. What is the resultant force on the 1.0C charge due to the other two

Answer 1

We know, force is given by :

$F=\dfrac{kq_1q_2}{r^2}$

Here, k in a constant.

$Net \ Force = F_1+F_2\\\\Net \ Force = k(\dfrac{Qq_1}{r_1^2}+\dfrac{Qq_2}{r_2^2})\\\\Net \ Force = k( \dfrac{-3}{50}+\dfrac{2}{50})\\\\Net \ Force = 9\times 10^9\times \dfrac{-1}{50}\ N\\\\Net\ Force = -1.8\times 10^8\ N$

Hence, this is the required solution.