Question

Three charges: q -2.00 x 10° C q2--3.20 x 10 C, and q-+ 1.50x 1o C are located along the x-axis at the following positions: x, -4.00m, x2-200m and x3 8.00 m. What is the magnitude and direction of the net force on q2 due to the other charges, qi and qs? 2-3.20 HC X3- 8.00 m x1-4.00m x2-2.00 m direction

Answer 1

since q1 and q2 are bot negative charge so q2 is repelled by q1. so the direction of force on it will be towards right.

Since q2 is negative charge and q3 is positive charge so q3 attract q2. Thus, the direction of force on it will be towards right.

Thus total net force direction on q2 will be towards right.

F_net=kq1*q3 /(x2-x1)^2 + kq2*q3/(x3-x2)^2

$F_{net}=\dfrac{9\times10^9\times2\times10^{-6}\times3.2\times10^{-6}}{(2-(-4))^2}+\dfrac{9\times10^9\times1.5\times10^{-6}\times3.2\times10^{-6}}{(8-2)^2}$

$F_{net}=1.6\times10^{-3}+1.2\times10^{-3}=2.8\times10^{-3}\, N$    towards right