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Speed of a charge at infinity

  • Thread starter alyssa.d
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Homework Statement


A charge of 3.40μC is held fixed at the origin. A second charge of 3.40μC is released from rest at the position (1.25m, 0.570m). If the mass of the second charge is 2.49 m, and its speed at infinity is 7.79 m/s, at what distance from the origin does the second charge attain half the speed it will have at infinity?


Homework Equations


I really am unsure which equations to use. However, I think that, when the second charge attains half the speed it will have at infinity, half its energy will be kinetic and half its energy will be potential. (Right? No? Maybe?) So some relevant equations would be:
Ki + Ui = Kf + Uf
F = kQ/r^2
ΔK = 1/2mvf^2 - 1/2mvi^2
ΔV = qΔV

You are solving for r.

The Attempt at a Solution


My issue in physics always is that I'm unsure how to set up the problem, and I never know which equations to pick. Lately this has been compounded by the fact that I've lost class and study time due to severe migraines, so I've been teaching myself almost everything (and this part of physics is something I've been struggling to pick up on my own).

Anyway, my attempt at a solution was this:
1/2mvi^2 + qVi = 1/2mvf^2 + qVf
1/2mvi^2 + q(kq/ri^2) = 1/2mf^2 + q(kq/rf^2)

I plugged in the numbers given in the problem (using 1.37 for ri, calculated from the initial position of the second charge). I got an rf value of 5.37. Any help would be appreciated!
 

Answers and Replies

  • #2
SammyS
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Homework Statement


A charge of 3.40μC is held fixed at the origin. A second charge of 3.40μC is released from rest at the position (1.25m, 0.570m). If the mass of the second charge is 2.49 m, and its speed at infinity is 7.79 m/s, at what distance from the origin does the second charge attain half the speed it will have at infinity?

Homework Equations


I really am unsure which equations to use. However, I think that, when the second charge attains half the speed it will have at infinity, half its energy will be kinetic and half its energy will be potential. (Right? No? Maybe?) So some relevant equations would be:
Ki + Ui = Kf + Uf
F = kQ/r^2
ΔK = 1/2mvf^2 - 1/2mvi^2
ΔV = qΔV

You are solving for r.

The Attempt at a Solution


My issue in physics always is that I'm unsure how to set up the problem, and I never know which equations to pick. Lately this has been compounded by the fact that I've lost class and study time due to severe migraines, so I've been teaching myself almost everything (and this part of physics is something I've been struggling to pick up on my own).

Anyway, my attempt at a solution was this:
1/2mvi^2 + qVi = 1/2mvf^2 + qVf
1/2mvi^2 + q(kq/ri^2) = 1/2mf^2 + q(kq/rf^2)

I plugged in the numbers given in the problem (using 1.37 for ri, calculated from the initial position of the second charge). I got an rf value of 5.37. Any help would be appreciated!
Hello alyssa. Welcome to PF !

(Use the X2 icon for superscripts, X2 for subscripts.)

Is "(Right? No? Maybe?)" a multiple choice? ... then No.

If the second charge has half it's max speed, does that mean it has half it's max Kinetic Energy? ... No.

You'll need to show what you did in more detail.

What's the KE at infinity?

What's the KE when the second charge is at half speed?
 
  • #3
Simon Bridge
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I think that, when the second charge attains half the speed it will have at infinity, half its energy will be kinetic and half its energy will be potential.
Conservation of energy would be a good place to start.

My issue in physics always is that I'm unsure how to set up the problem, and I never know which equations to pick.
Don't worry about remembering the equations to pick - sketch a diagram of the situation. This is the thing to practice.

The free charge will accelerate radially away from the fixed one ... so you just need to know the initial radius and the angle rather than the x-y components.

Anyway, my attempt at a solution was this:
1/2mvi^2 + qVi = 1/2mvf^2 + qVf
1/2mvi^2 + q(kq/ri^2) = 1/2mf^2 + q(kq/rf^2)

I plugged in the numbers given in the problem (using 1.37 for ri, calculated from the initial position of the second charge). I got an rf value of 5.37. Any help would be appreciated!
What is the initial velocity of the free charge?

Try thinking about this in terms of work. dW=F(r).d
Do you know the equation for the potential of a point charge?
 
  • #4
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Yes. The initial velocity of the free charge is 0, since it's released from rest. The equation for the potential of a point charge is V = kq/r. I feel like I am making this more difficult than it needs to be, like I am missing something very simple just in terms of how I'm thinking about setting it up.
 
  • #5
SammyS
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Yes. The initial velocity of the free charge is 0, since it's released from rest. The equation for the potential of a point charge is V = kq/r. I feel like I am making this more difficult than it needs to be, like I am missing something very simple just in terms of how I'm thinking about setting it up.
The problem appears to me to be over-determined.

ri is given, so you have initial potential energy, Ui = q1V1 = kq2/r1.

Ki = 0.

Uf = 0, so you can find Kf from conservation of energy.

You are given the mass of the charged particle, so you can determine the speed at infinity.

Does that agree with the speed given?

Well, if that mass is 2.49 grams, then yes, the speed at infinity is consistent with the other given information.
 
  • #6
Simon Bridge
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If I use K for kinetic energy:

* You have ##v_\infty = \sqrt{2K_\infty/m}##

* You know that loss in potential energy moving to position r = gain in kinetic energy.

* The speed at position r must be given by ##v=\sqrt{2K(r)/m}##

* You want to find r so that ##v(r)/v_\infty=1/2##.

That help?
 

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