Question

Two small spheres, each carrying a net positive charge, are separated by 0.400 m. You have been asked to perform measurements that will allow you to determine the charge on each sphere. You set up a coordinate system with one sphere (charge q1) at the origin and the other sphere (charge q2) at x = +0.400 m. Available to you are a third sphere with net charge q3 = 4.00 x 10-6 C and an apparatus that can accurately measure the location of this sphere and the net force on it. First you place the third sphere on the x-axis at x = 0.200 m; you measure the net force on it to be 4.50 N in the +x-direction. Then you move the third sphere to x = +0.600 m and measure the net force on it now to be 3.50 N in the +x-direction. (a) Calculate q1 and q2. (b) What is the net force (magnitude and direction) on q3 if it is placed on the x-axis at x = -0.200 m? (c) At what value of x (other than x = {q) could q3 be placed so that the net force on it is zero?

Short answer

(a) the value of \({q_1}\) is \(8.0\mu C\), and that of \(q\mu C\)is \(3.0\mu C\)

(b) The net force on \({q_3}\) if it is placed on the \(x - axis{\rm{ }} = - 0.200m\)is \( - 7.50N\)

(c) At \(x = + 0.248m\),\({q_3}\)can be placed so that the net force is zero.

Step-by-step solution

Figure representing first case

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:

Calculating \({q_1}\)and \({q_2}\)for the first case:

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: \({q_1} - {q_2} = 5.0 \times {10^{ - 6}}\)

Figure representation of second and third case

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 \( - 7.50N\)

Calculating the distance

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

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 \( + 0.248m\)