91Ó°ÊÓ

Given:

\(\begin{aligned}{x_1} = 0m\\{x_2} = + 0.4m\\{q_3} = 4.0 \times {10^{ - 6}}C\\{x_{3,a}} = + 0.20m\\\sum {{F_{3,a}} = + 4.50N} \\{x_{3,b}} = + 0.60m\\\sum {{F_{3,b}} = + 3.50N} \\{x_{3,c}} = - 0.20m\\\sum {{F_{3,c}} = 0N} \\k = 9.0 \times {10^9}N.{m^2}/{C^2}\\\end{aligned}\)

First, we need to draw the figure that represents each case. And then we need to draw the forces exerted on the third sphere by the two other spheres. Since we already know that like charges repel and that is why we draw the force exerted on the third sphere as shown below.

For the first case: The first sphere exerts a force on the third sphere to the right and the second ball exerts a force to the left, therefore the resultant figure is:

Applying Newtons second law of motion when the third sphere is at \(x = {\rm{ }} + 0.20m\)

\(\sum {{F_{3,a}} = {F_1} - {F_2}} \)

Now, we know that the electric force between any two charge is given by:

\(F = \frac{{k{q_1}{q_2}}}{{{r^2}}}\)

Therefore: \(\sum {{F_{3,a}} = \frac{{k\left| {{q_1}{q_3}} \right|}}{{{r_{13}}^3}} - \frac{{k\left| {{q_2}{q_3}} \right|}}{{{r_{23}}^2}}} \)

Solving it we get: \({q_1} - {q_2} = 5.0 \times {10^{ - 6}}\)

For the second case:

Figure representation of third case:

Now using the same approach, we find:

\(\begin{aligned}\sum {{F_{3,c}} = - {F_1} - {F_2}} \\\sum {{F_{3,c}}} = - (\frac{{k\left| {{q_1}{q_3}} \right|}}{{{r_{13}}^2}} + \frac{{k\left| {{q_2}{q_3}} \right|}}{{{r_{23}}^2}})\end{aligned}\)

Putting the values, we get:

\(\sum {{F_{3,c}} = - 7.50N} \)

Therefore, the net force on \({q_3}\) is \( - 7.50N\)

\(\sum {{F_3} = {F_1} - {F_2} = 0} \)

From the above approach we get: \(0 = \frac{{{q_1}}}{{{{({x_3})}^2}}} - \frac{{{q_2}}}{{{{(0.40 - {x_3})}^2}}}\)

Putting the values of \({q_1}\) and \({q_2}\)we get:

\({x_3} = 0.248m\)Therefore, the distance of \({x_3}\) is \( + 0.248m\)