Finding Distance Using Coulomb's Law

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Homework Help Overview

The discussion revolves around a problem involving Coulomb's Law and the forces acting on a charged object on a horizontal plane. Participants explore the relationship between electric force and friction, as well as the implications of energy conservation in determining the final position of the charged object.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss setting the electric force equal to the force of friction and question the implications of this balance. There is exploration of using conservation of energy, with references to kinetic and potential energy, and the work done by friction. Some participants express confusion about the initial conditions and how they affect the calculations.

Discussion Status

The discussion is active, with participants providing guidance on the use of energy conservation and the correct interpretation of forces. There is acknowledgment of the need to clarify the work done by friction and its relationship to the initial distance. Multiple interpretations of the problem are being explored, particularly regarding the setup and equations involved.

Contextual Notes

Participants note that the problem involves a fixed charge and a moving charge, with specific constraints such as the coefficient of friction and the initial distance. There is mention of the potential complexity of the solution, which may not align with expectations from the professor.

thetest
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Homework Statement


Homework Equations


The Attempt at a Solution



My reasoning is that when the charge stops that means that the force of friction equals the electric force. So I find the distance by setting the two equations equal to each other and solve for x which gives (keq1q2/(mgk))^(1/2), however what about the initial distance that they were set at? Do I add it to this value or subtract it or is it already included within the value of x that I solved for? Any help would be greatly appreciated. Thanks!
 
Last edited:
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thetest said:

Homework Statement


Charge q1>0 is fixed on the surface of the horizontal plane as shown on the picture. An object of mass m, having charge q2>0 is initially placed at a distance l0 from the first charge. Find final position of this object on the plane once it is released. Friction coefficient between object and the surface is k.


Homework Equations


Ff = kmg; Fe = kq1q2/R^2


The Attempt at a Solution



My reasoning is that when the charge stops that means that the force of friction equals the electric force.

That won't work. When the force of friction equals the electric force, the net force is 0, so the object will keep on moving.
Use conservation of energy. The energy to move an object a certain distance with a constant friction is easy to find.
 
willem2 said:
That won't work. When the force of friction equals the electric force, the net force is 0, so the object will keep on moving.
Use conservation of energy. The energy to move an object a certain distance with a constant friction is easy to find.

How can we use the conservation of energy? Are you talking about the equation K1+U1+Wother= K2 + U2? But even if we use that we are not provided a velocity nor does the charge change in height so we can't use potential?
 
thetest said:
How can we use the conservation of energy? Are you talking about the equation K1+U1+Wother= K2 + U2? But even if we use that we are not provided a velocity nor does the charge change in height so we can't use potential?

You don't need a velocity. The velocity of the charge is 0 at the start and at the end, so the initial and final kinetic energies are 0. This leaves only the electrical potential energy and the work that friction does on the charge.
 
willem2 said:
You don't need a velocity. The velocity of the charge is 0 at the start and at the end, so the initial and final kinetic energies are 0. This leaves only the electrical potential energy and the work that friction does on the charge.

Do we solve for x
 
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thetest said:
Oh okay so we have U1 + Wother = U2; keq1q2/l0 - kmg = kq1q2/x. Then we solve for x?

kmg is the force of friction, not the work done
 
willem2 said:
kmg is the force of friction, not the work done
Then we solve for x to find the distance right?
 
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thetest said:
Oh I'm sorry you're right. W = Ff*d; so then it would be U1 + Wother = U2; keq1q2/l0 - kmgx = kq1q2/x. Then we solve for x to find the distance right?

Yes! Don't confuse the different k's tough.
 
willem2 said:
Yes! Don't confuse the different k's tough.

Awesome thank you! I do have a question though, when I solve for x, I get xq1q2/l0 - (4*pi*e0)(kmgx^2) = q1q2. Is there a way to simplify this?
 
  • #10
The work done by friction should be [itex]kmg(x - l_0)[/itex] since the charge starts at [itex]l_0[/itex]

You get a quadratic equation
 
  • #11
willem2 said:
The work done by friction should be [itex]kmg(x - l_0)[/itex] since the charge starts at [itex]l_0[/itex]

You get a quadratic equation

Oh okay and then we use the quadratic equation to solve for x? It is just the + value right?
 
  • #12
thetest said:
Oh okay and then we use the quadratic equation to solve for x? It is just the + value right?

I'm not sure. x should be positive and there should be only 1 positive solution.
 
  • #13
willem2 said:
I'm not sure. x should be positive and there should be only 1 positive solution.

Oh okay I found the positive solution. Are you sure this is the way to solve the problem because my professor says that the answers should not be very complicated? Thanks again though!
 

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