# Homework Help: Charges and Coulomb's Law

1. Jan 30, 2010

### electroguy02

1. The problem statement, all variables and given/known data

Two positive point charges q are placed on the y-axis at a and -a. A negative point charge -Q is located at some point x on the +x-axis.

Find the x-component of the net force that the two positive charges exert on -Q.

Express your answer in terms of the variables q, Q, x, and a, with any necessary constants.

2. Relevant equations

Coulomb's Law:

F = kq_1q_2/r^2

3. The attempt at a solution

The upper particle on the y-axis is q_1, the lower particle on the y-axis is q_2, and the particle on the x-axis is q_3.

q_1 and q_2 have the same "pull" on q_3, so the vertical y-components of the force cancels out. Also, the force due to q_1 on q_3 is equal to the pull that q_2 has on q_3. The force on q_3 is equal to:

F_q_3 = F_13 + F_23

F_13 = kq_1q_3/r^2

k = 8.99 * 10^9
q_1 = q
q_3 = -Q
r^2 = a^2 + x^2 (pythagorean thereom to find the distance from q_1 to q_3)

The magnitude of the force F_13 (particle one on particle three) is:

F_13 = (8.99 * 10^9)(q)(-Q)/(a^2 + x^2)

The x-component of this force is

F_13x = (8.99 * 10^9)(q)(-Q)/(a^2 + x^2) * x/(sqrt(x^2 + a^2))

Since F_q_3 = F_13 + F_23, and F_13 = F_23,

F_q_3x = 2[(8.99 * 10^9)(q)(-Q)/(a^2 + x^2) * x/(sqrt(x^2 + a^2))]

This is also what my teacher got for this problem, but it's incorrect. Can somebody please help me point out what the problem is?

F_q_3x = 2[(8.99 * 10^9)(q)(Q)/(a^2 + x^2) * x/(sqrt(x^2 + a^2))]
F_q_3x = 2[(q)(Q)/(a^2 + x^2) * x/(sqrt(x^2 + a^2))]
F_q_3x = 2[(q)(Q)x/(a^2 + x^2)^(3/2)]

2. Jan 31, 2010

### electroguy02

I don't understand.

Since I'm looking for the x-component of the force, I'd need the force directed along the x-axis, right? If the "height" of the triangle of one force is a, and the "base" is x, then the hypotenuse of the triangle would be sqrt(a^2 + x^2), right?

Why would I need the vertical a value to solve for the x-component, especially when the vertical components cancel out?

3. Jan 31, 2010

### rl.bhat

Yes. You are right. Your answer is correct. Probably you have to include the sign, because the force in pointing towards the -ve x axis.

4. Jan 31, 2010

### electroguy02

Argh... I figured out what I did "wrong" in the end.

I put that k = 8.99 * 10^9, so instead of writing down k, I wrote down the numerical value. And the system kept telling me that it was wrong. So the answer was:

F_q_3x = 2[(k)(q)(-Q)/(a^2 + x^2) * x/(sqrt(x^2 + a^2))]