Electric force problem (determining unknown position to make zero net force)

AI Thread Summary
To achieve zero net force on a proton placed between a +2.0-nC charge and a -4.0-nC charge, it must be positioned outside the interval defined by these charges. The calculations using the equation F01 = F02 yield two potential solutions for the distance (r): +0.414 cm and -2.41 cm. The positive solution indicates a position where the forces would act in the same direction, which does not satisfy the condition for zero net force. Therefore, the correct placement for the proton is at -2.41 cm, to the left of the positive charge, where the forces exerted by both charges will balance each other out. Understanding the direction of forces is crucial in determining the appropriate position for achieving equilibrium.
KYUI
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Homework Statement


A +2.0-nC charge is at the origin and a -4.0-nC charge is at x = 1.0-cm.
At what x-coordinate could you place a proton so that is would experience no net force?

[NOTE by me] I put q1 as the positive charge (+2.0nC) and q2 as the negative charge (-4.0nC). q0, of course, is the proton.

Homework Equations


- quadratic formula
- F0 = F01 - F02 which becomes F01 = F02 when I replace F0 with zero (no net force).

The Attempt at a Solution


So I know that the proton must be either to the left of the positive charge or the right of the negative charge. It can't be in between these two charges for it to have zero net force.

I plugged in numbers into this equation (F01 = F02) which became:
(2 x 10-9)/(r2) = (4 x 10-9) / ((1.0 x 10-2) + r)2
I simplified this and got:
0 = r2 + (2 x 10-2)r - (1 x 10-4)

Then I solved for r using quadratic formula and got:
r = (+ 4.1421 x 10-3) or (- 2.4142 x 10-2).

I thought the negative number would automatically cancel out because r is distance but the actual answer goes with the negative number. I'm confused at how cramster.com explained it.

The book is called Physics for Scientists and Engineers A Strategic Approach (2nd) by Randall Dewey Knight) and its chapter 26 number 47. Here's a direct link (have to be a member though; it's free to look at odd numbers): http://physics.cramster.com/physics-for-scientists-and-engineers-a-strategic-approach-2nd-problem-9-718769.aspx

Thanks.
 
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Well r is measured relative to the origin, so if you get r is positive, you place the proton where the force between the proton and the +ve charge pushes it towards the negative charge.

Otherwise, r is negative, and the proton is placed to the left of the +ve charge. The force between the proton and the positve charge pushes it away i.e. towards the left. The Force between the proton and the -ve charge will attract it i.e. attract it towards the right. Thus with r negative, there exists a location where the net force is zero.

Do you understand a little more now?
 
Thanks for answering my question. I really appreciate it. And I'm sorry I still don't understand :(

I understand that r can be negative and relation to the origin but I don't understand how to determine which number (that I found for r) is used as the answer.

If r is positive, the proton will be on left side of the positive charge, right? I thought it would be attracted to the negative charge and repelled by positive charge to possibly get a net zero force, no?

And same thing for negative r. It will be on the right side of the negative charge, right? Then it would be attracted to the negative charge and repelled by positive charge in the same way to possibly get a net zero force.

Mm I don't know if I'm making my question clear. Do you have to determine the position by calculating it and prove if the net force is zero or not?

Thanks again.
 
KYUI said:
Thanks for answering my question. I really appreciate it. And I'm sorry I still don't understand :(

I understand that r can be negative and relation to the origin but I don't understand how to determine which number (that I found for r) is used as the answer.

If r is positive, the proton will be on left side of the positive charge, right? I thought it would be attracted to the negative charge and repelled by positive charge to possibly get a net zero force, no?

And same thing for negative r. It will be on the right side of the negative charge, right? Then it would be attracted to the negative charge and repelled by positive charge in the same way to possibly get a net zero force.

Mm I don't know if I'm making my question clear. Do you have to determine the position by calculating it and prove if the net force is zero or not?

Thanks again.

Since you used the condition F_{01} = F_{02}, whatever value of r you get, that is where the net force is zero. But I can't see how the book explained it as I need to join. I understand what you're saying and both answers should be correct.
 
This is what it said:

"x = +0.414cm or -2.41cm at the above two points the magnitudes of the two forces are equal. At point .414cm the magnitudes of both the forces are equal but the directions are same. But we require the forces to be in opposite direction. So the required position of the proton is 2.41cms on the side opposite q_2. "

They also used same equation as me: F01 = F02.
 
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