Question

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.

Answer 1

Answer:

Q₂=-3.01 μC

Step-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:

$F=k_{e}*\frac{Q_{1}*Q_{2}}{r^{2}}$

Where ke is Coulomb's constant: $k_{e}=9*10^{9} \frac{N*m^{2}}{C^{2}}$

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:

$Q_{1}=5*10^{-6}C (x=0m)$

$Q_{2}=? (x=0.12m)$

$Q_{3}=2*10^{-6}C (first,x=0.24m; then, x=0.5325m)$

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):

$F_{23}=9*10^{9} \frac{N*m^{2}}{C^{2}}*\frac{Q_{2}*Q_{3}}{(r_{23})^{2}}$

$F_{23}=9*10^{9} \frac{N*m^{2}}{C^{2}}*\frac{Q_{2}*2*10^{-6}C}{(0.12m)^{2}}$

$F_{23}=1250000\frac{N}{C} *Q_{2}$

Now, for the repelling force of Q₁ on Q₃ (F₁₃):

$F_{13}=9*10^{9} \frac{N*m^{2}}{C^{2}}*\frac{Q_{1}*Q_{3}}{(r_{13})^{2}}$

$F_{13}=9*10^{9} \frac{N*m^{2}}{C^{2}}*\frac{5*10^{-6}C*2*10^{-6}C}{(0.24m)^{2}}$

$F_{13}=1.5625N$

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:

$2.2N=F_{23}-F_{13}\\\\2.2N=1250000\frac{N}{C} *Q_{2}-1.5625N\\1250000\frac{N}{C} *Q_{2}=2.2N+1.5625N\\1250000\frac{N}{C} *Q_{2}=3.7625N\\Q_{2}=3.01*10^{-6}C$

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.

Answer 2

The value of Q2 = 3.01 *$10^{-6}$ C

What is Coulomb's law?

According to Coulomb, the electric force for charges at rest has the following properties: Like charges repel each other; unlike charges attract.

Q₁ = 5μC (x = 0)

Q₂ = ? (x = 12cm)

Q₃ = 2μC

Using Coulomb's law,

F=  Ke* Q1* Q2 /r²

where, Ke= 9* $10^{9}$ Nm²/ C²

Q1=  5*$10^{-6}$

Q2 = ?

Q3= 2*$10^{-6}$

Now,

F23=  9* $10^{9}$ * Q2 * 2*$10^{-6}$ / (0.12)²

F23= 1250000 / Q2

Now, F13 = 9* $10^{9}$ *  5*$10^{-6}$ * 2*$10^{-6}$ / (0.24)²

F13= 1.5625 N

As stated , 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.

2.2= F23 - F13

2.2 =  1250000 / Q2 - 1.5625 N

1250000 / Q2 = 3.7625

Q2 = 3.01 *$10^{-6}$ C

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