Acharge 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.

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:

Where ke is Coulomb's constant:

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:

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):

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

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:

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.

step-by-step explanation:

equation

you are asked to write the equation

  expression = 0

using the given expression. when you do that, you have

  2x^2-2x-12 = 0

factoring

you can see that each of the coefficients is even, so a factor of 2 can be removed immediately.

  2(x^2 -x -6) = 0

if you multiply the binomials (x+a)(x+b), you get

  (x +a)(x +b) = x^2 +(a+b)x +ab

this tells you the coefficient of x is the sum of two factors of the constant term. here, that means -1 is the sum of two factors of -6. we want to know those factors so we can factor the trinomial.

  -6 = -1×6 = -2×3 = -3×2 = -6×1

the factor pair that sums to -1 is {-3, 2}, so our factorization is

  2(x -3)(x +2) = 0

solutions

the solutions to this equation are the values of x that make the product zero. a product is zero only when only of the factors of the product is zero. here, that means

  x - 3 = 0

  x = 3 . . . add 3 to both sides of the equation

or

  x +2 = 0

  x = -2 . . . add -2 to both sides of the equation

the solutions to the equation are x=3 or x=-2. both of these values of x satisfy the equation—they make it be true.

step-by-step explanation: