Finding charge q2 from net force on particle 2 as particle 3 moves along x axis

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lydia_y620
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Homework Statement


Figure (a) shows charged particles 1 and 2 that are fixed in place on an x axis. Particle 1 has a charge with a magnitude of |q1| = 19e. Particle 3 of charge q3 = +16e is initially on the x axis near particle 2.Then particle 3 is gradually moved in the positive direction of the x axis. As a result, the magnitude of the net electrostatic force on particle 2 due to particles 1 and 3 changes. Figure (b) gives the x component of that net force as a function of the position x of particle 3. The scale of the x axis is set by xs = 1.70 m. The plot has an asymptote of F2,net = 0.8688 × 10-25 N as x → ∞. As a multiple of e and including the sign, what is the charge q2 of particle 2?
upload_2017-1-31_1-36-40.png


Homework Equations


F = kq1q2/r2

The Attempt at a Solution


As x --> ∞, the force on particle 2 comes just from particle 1, so the force from particle 1 is always 0.8688 x 10^-25 N because particles 1 and 2 are stationary.
When x = 0.85 for particle 3, the net force on particle 2 is 0, so the force from particle 1 must equal the force from particle 3.
This means that 0.8688 x 10^-25 = k(q2*16e)/0.85^2
Is this correct? Apparently my answer is incorrect after I tried solving for q2 in terms of e.
 
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lydia_y620 said:
1. Figure (a) shows charged particles 1 and 2 that are fixed in place on an x axis. Particle 1 has a charge with a magnitude of |q1| = 19e. Particle 3 of charge q3 = +16e is initially on the x axis near particle 2.Then particle 3 is gradually moved in the positive direction of the x axis. As a result, the magnitude of the net electrostatic force on particle 2 due to particles 1 and 3 changes. Figure (b) gives the x component of that net force as a function of the position x of particle 3. The scale of the x axis is set by xs = 1.70 m. The plot has an asymptote of F2,net = 0.8688 × 10-25 N as x → ∞. As a multiple of e and including the sign, what is the charge q2 of particle 2?
View attachment 112319

Homework Equations


F = kq1q2/r2

The Attempt at a Solution


As x --> ∞, the force on particle 2 comes just from particle 1, so the force from particle 1 is always 0.8688 x 10^-25 N because particles 1 and 2 are stationary.
When x = 0.85 for particle 3, the net force on particle 2 is 0, so the force from particle 1 must equal the force from particle 3.
This means that 0.8688 x 10^-25 = k(q2*16e)/0.85^2
Is this correct? Apparently my answer is incorrect after I tried solving for q2 in terms of e.[/B]
Please post your final answer.
I note that it asks for the charge as a multiple of e, but your last equation would naturally give a multiple of 1/e.
 
haruspex said:
Please post your final answer.
I note that it asks for the charge as a multiple of e, but your last equation would naturally give a multiple of 1/e.
My final answer was (4.359 x 10^-37)/e, which, like you've said, doesn't make sense because it's a multiple of 1/e.
 
lydia_y620 said:
My final answer was (4.359 x 10^-37)/e, which, like you've said, doesn't make sense because it's a multiple of 1/e.
It does make sense. It arises naturally from q1=Fr2/(kq2).
So express it as constant x e instead. e is a known value.