Three charges are placed as shown in Fig. P21.95. The magnitude of q1 is 2.00 mC, but its sign and the value of the charge q2 are not known. Charge q3 is +4.00 mC, and the net force F on q3 is entirely in the negative x-direction. (a) Considering the different possible signs of q1, there are four possible force diagrams representing the forces F 1 and F2 that q1 and q2 exert on q3. Sketch these four possible force configurations. (b) Using the sketches from part (a) and the direction of F , deduce the signs of the charges q1 and q2. (c) Calculate the magnitude of q2. (d) Determine F, the magnitude of the net force on q3

Above are all the possible outcomes.

Step2: All the possible components

To find the signs we need to guess the possible components and compare it with the net force direction

Case1:

Vertical component: There should be a vertical component and have a value in upward direction.

Horizontal component: May exist in any one and the component is larger than other

Comparing with the previous prediction we get the case it not right as the net force has only vertical component to left.

Case2:

Vertical component: There should be a vertical component and have a value in downward direction.

Horizontal component: May exist in any one and the component is larger than other

Comparing with the previous prediction we get the case is not correct as the net force has only horizontal component to left.

Case3:

Vertical component: may exist in case of one vertical component is larger than other

Horizontal component: Must exist with value in left direction

Comparing with the previous prediction we get the case is possibly right as the net force has only horizontal component to left.

Case4:

Vertical component: may exist in case of one vertical component is larger than other

Horizontal component: Must exist with value in right direction

Comparing with the previous prediction we get the case is not right as the net force has only horizontal component to left.

Therefore, we conclude that $q_1$and should be negative and $q_2$should be positive.

The angle ${\theta }_1$and 54 N horizontal is:

$$\begin{gathered} {\theta }_1=\mathrm{arcco}s\left(\frac{4}{5}\right) \\ =36.86° \end{gathered}$$

${\theta }_2$and$F_2$is:

$$\begin{gathered} {\theta }_2=\mathrm{arcco}s\left(\frac{3}{5}\right) \\ =53.13° \end{gathered}$$

Now, the vertical component must be zero so:

Now force $F_1$is give by:

$$\begin{gathered} F_1=\frac{k{q}_1{q}_3}{{\left(r_{13}\right)}^2}=\frac{9\times {10}^9\times 2.00\times {10}^{-6}\times 14.00\times {10}^{-6}}{{\left(0.4\right)}^2} \\ =45N \end{gathered}$$

Force $F_2$is: 33.8N

Substituting the value, we get: $q_2=8.45\times {10}^{-7}C$

The magnitude of the force F is =

$$\begin{gathered} F=F_1\cos \left(36.86\right)+F_2\cos \left(53.13\right) \\ =45\cos \left(36.86\right)+33.8\cos \left(53.13\right) \\ =54N \end{gathered}$$

Therefore, the force is 54N