Question

Coulomb's law for the magnitude of the force F between two particles with charges Q and Q′ separated by a distance d is |F|=K|QQ′|/d2, where K=14πϵ0, and ϵ0=8.854×10^−12C^2/(N⋅m2) is the permittivity of free space. Consider two point charges located on the x axis: one charge, q1 = -11.5 nC , is located at x1 = -1.655 m ; the second charge, q2 = 38.0 nC , is at the origin. What is the net force exerted by these two charges on a third charge q3 = 49.5 nC placed between q1 and q2 at x3 = -1.170 m ?

Answer 1

The third charge $q_3=49.5\,nC$ is placed to the right at a distance of $x_{31}=1.655-1.170=0.485\,m$ of charge $q_1=-11.5\,nC$ and to the left at a distance of $x_{32}=1.170\,m$ of charge $q_2=38\,nC$

Force on charge $q_3$ due to charge $q_1$ is directed towards left (towards negative X-direction)

Force on charge  $q_3$ due to charge $q_2$ is also directed towards left (towards negative X-direction)

Magnitude of net force on charge $q_3$ ,

$F_3=\frac{q_3q_1}{4\pi\epsilon_0 x_{31}^2}+\frac{q_3q_2}{4\pi\epsilon_0 x_{32}^2}$

$F_3=\frac{8.99*10^{9}*49.5*10^{-9}*11.5*10^{-9}}{(0.485)^2}+\frac{8.99*10^{9}*49.5*10^{-9}*38.0*10^{-9}}{(1.170)^2}$

$F_3=0.0000341\,N=3.41*10^{-5}\,N$ to the left