We have
We have
Q = magnitude of charge at each of the three locations A, B and C=3.5 x 10⁻⁶ C
r₁ = distance of charge at origin (A) from charge at B = 6cm = 0.06 m
r₂ = distance of charge at origin from charge at C = 7cm=0.07m
F₁ = magnitude of force by charge at B on charge at origin
F₂ = magnitude of force by charge at C on charge at origin
Let's calculate F₁
[tex]F_1=\frac{kQ^2}{r^2_1}[/tex]k is a constant that is 9x10^9
we substitute the values
[tex]F_1=\frac{(9\times10^9)(3.5\times10^{-6})^2}{0.06^2}[/tex][tex]F_1=30.625N_{}[/tex]then for the force with charge C
[tex]F_2_{}=\frac{kQ^2}{r^2_2}[/tex][tex]F_1=\frac{(9\times10^9)(3.5\times10^{-6})^2}{0.07^2}=22.5N[/tex]Then we calculate the net force
[tex]F=F_1+F_2[/tex][tex]F=30.625+22.5[/tex][tex]F=53.125N[/tex]As we can see in the diagram the forces go to the left therefore the solution is
[tex]F=-53.1N[/tex]
