- #1
electroguy02
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Homework Statement
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.
Homework Equations
Coulomb's Law:
F = kq_1q_2/r^2
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?
Thanks in advance.
ADD: I've also tried:
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)]