Coulomb's Law problem: Charged particles and the net electric field and force

[lydia_y620]

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?

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.

 

[haruspex]

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.

 

[lydia_y620]

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.

 

[haruspex]

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.

 

[NSchool2000]

The checker is not taking it as that.

 

[haruspex]

The checker is not taking it as that.

What checker is not taking what as what?