Q:

A charge of 5.0 mu or micro CC is located at x = 0, y = 0 and a charge Q2 is located at x = 12 cm, y = 0. The force on a 2 mu or micro CC charge at x = 24 cm, y = 0 is 2.2 N, pointing in the negative x direction. When this 2 mu or micro CC charge is positioned at x = 53.25 cm, y = 0, the force on it is zero. Determine the charge Q2.

Accepted Solution

A:
Answer:Q₂=-3.01 μCStep-by-step explanation:First of all, since no charge has a "y" coordinate, we can just focus on the presented "x" coordinates. Also we should remember that micro C is denoted as μC. We can then list the charges that we have, and their respective "x" coordinates:Q₁ = 5μC (x = 0)Q₂ = ? (x = 12cm)Q₃ = 2μC (first x = 24cm, and then x = 53.25cm)So the first major piece of information we are given is that the force on a 2μC charge (our Q₃) at x = 24 cm is 2.2 N, pointing in the negative x direction. What it is saying is that, basically, there is a negative charge somewhere that is pulling our Q₃ back into the negative x direction. This is because opposite charges attract each other while similar charges repel each other. In this sense, given that Q₃ is positive, and that Q₁, which is positive as well, must be repelling it to the positive x direction, there should be a negative charge that is strong enough to pull it back, despite the Q₁ repulsion on Q₃. This negative charge is Q₂.Please refer to the image attached for a clearer view on the positions of each charge and the forces exerted on Q₃.Now, we should apply Coulomb's Law, which is used to determine the exerted force between charges:[tex]F=k_{e}*\frac{Q_{1}*Q_{2}}{r^{2}}[/tex]Where ke is Coulomb's constant: [tex]k_{e}=9*10^{9} \frac{N*m^{2}}{C^{2}}[/tex]And r is the distance between charges, in m.But, before continuing, let us transform μC into C, and cm into m, so we can use these numbers in Coulomb's Law:[tex]Q_{1}=5*10^{-6}C (x=0m)[/tex][tex]Q_{2}=? (x=0.12m)[/tex][tex]Q_{3}=2*10^{-6}C (first,x=0.24m; then, x=0.5325m)[/tex]Now, remember that, as mentioned before, Q₁ induces a repelling force on Q₃, while Q₂ induces an attraction force on Q₃. This means that we have to calculate the force of Q₂ on Q₃ (F₂₃) and then subtract the force of Q₁ on Q₃ (F₁₃). To do this, we have to first determine the distance (r) between the charges at play:In the first scenario, Q₃ is in x = 0.24m and Q₂ is in x = 0.12m. The distance between them is therefore r₂₃ = 0.24m - 0.12m = 0.12m. On the other hand, the distance between Q₁ and Q₃ is r₁₃ = 0.24m - 0 = 0.24m. Let us proceed with Coulomb's Law for the attractive force of Q₂ on Q₃ (F₂₃):[tex]F_{23}=9*10^{9} \frac{N*m^{2}}{C^{2}}*\frac{Q_{2}*Q_{3}}{(r_{23})^{2}}[/tex][tex]F_{23}=9*10^{9} \frac{N*m^{2}}{C^{2}}*\frac{Q_{2}*2*10^{-6}C}{(0.12m)^{2}}[/tex][tex]F_{23}=1250000\frac{N}{C} *Q_{2}[/tex]Now, for the repelling force of Q₁ on Q₃ (F₁₃):[tex]F_{13}=9*10^{9} \frac{N*m^{2}}{C^{2}}*\frac{Q_{1}*Q_{3}}{(r_{13})^{2}}[/tex][tex]F_{13}=9*10^{9} \frac{N*m^{2}}{C^{2}}*\frac{5*10^{-6}C*2*10^{-6}C}{(0.24m)^{2}}[/tex][tex]F_{13}=1.5625N[/tex]As stated before, we should do the following subtraction in order to get our 2.2 N force that is exerted on Q₃, which is an attractive one because of the negative charge of Q₂:[tex]2.2N=F_{23}-F_{13}\\\\2.2N=1250000\frac{N}{C} *Q_{2}-1.5625N\\1250000\frac{N}{C} *Q_{2}=2.2N+1.5625N\\1250000\frac{N}{C} *Q_{2}=3.7625N\\Q_{2}=3.01*10^{-6}C[/tex]As a result, we got the absolute charge of Q₂, which is 3.01 μC; and since we know it is a negative charge, Q₂ is -3.01 μC.Notice how we didn't need the other scenario where our Q₃ is positioned at x = 0.5325m? This is because we had enough information in the first scenario to determine the charge of Q₂.