Finding the Distance: Charge Particle Motion

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SUMMARY

The discussion focuses on calculating the distance a charged particle travels before it stops and reverses direction due to the electrostatic force exerted by a fixed charge. A charge of -4.00 µC is fixed, while a particle with a mass of 2.50 x 10^-3 kg and a charge of -3.00 µC is fired towards it at an initial speed of 15.0 m/s from a distance of 55.0 cm. Participants emphasize using conservation of energy principles and Newton's second law to derive the correct equations, ultimately determining that the particle moves approximately 0.32 m before stopping.

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  • Understanding of electrostatic forces and Coulomb's law
  • Familiarity with Newton's second law of motion
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flower76
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Another problem I'm struggling with this one I don't have a clue where to start. Any hints to a starting point would be greatly appreciated.

A charge of -4.00uC is fixed in place. From a horizontal distance of 55.0cm a particle of mass 2.50 x 10^-3 kg and charge -3.00uC is fires with an intial speed of 15.0m/s directly toward the fixed charge. How far does the particle travel before it stops and begins to return back?
 
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The net force on the moving charge should be the force of repulsion it experiences. With this, you can set up an equation using Newton's 2nd Law.

Take note though, that the net force (and hence the acceleration) of the moving particle is not constant, as it varies with the distance between the two particles. Hence, you will need to express acceleration as \frac{dv}{dt}, or some other equivalent expression.

Once you have carried out integration, do not forget to add a constant of integration. You can solve for it by using the initial conditions provided in the quesion.
 
pizzasky said:
The net force on the moving charge should be the force of repulsion it experiences. With this, you can set up an equation using Newton's 2nd Law.

Take note though, that the net force (and hence the acceleration) of the moving particle is not constant, as it varies with the distance between the two particles. Hence, you will need to express acceleration as \frac{dv}{dt}, or some other equivalent expression.

Once you have carried out integration, do not forget to add a constant of integration. You can solve for it by using the initial conditions provided in the quesion.

You don't need to integrate. The basic idea is similar to the scattering experiment. The moving charge stops when all its kinetic energy has become zero. A simple energy conservation for the moving charge is to be done here. You need to figure out the distance of closest approach from the static charge.

Pizzasky's approach will land you with the same answer but the basic idea is energy conservation of the moving charge in classical terms.
 
So here's what I've come up with:

Q = fixed charge
q = moving charge

kQq/(x-0.55m) = kQq/r +(1/2mv^2)

Solving for x.

I'm not sure if x-0.55 is correct? Or even if I've set up the equation correctly?
 
It's almost right. What's "r"?
 
r is the initial distance between the charges = 0.55m
 
You're almost there. Using conservation of energy;

U_{final} = E_{k} + U_{intial}

\frac{kQq}{0.55 - x} = E_{k} + \frac{kQq}{0.55}

~H

Edit: Sorry Doc, guess I was typing for a long time
 
Last edited:
flower76 said:
r is the initial distance between the charges = 0.55m
Right. So, if the moving particle moves a distance x, what's the new separation distance between them?
 
Hootenanny said:
However, it would be easier to consider the change in distance;

\frac{kQq}{\Delta r} = E_{k}

Since \Delta r = 0.55 - x ;

\frac{kQq}{0.55 - x} = E_{k}
:eek: You might want to reconsider that, Hoot.
 
  • #10
I'm not sure, so if

\frac{kQq}{0.55 - x} = E_{k}

then Ek is the PE + KE of q?

So my equation would be:

\frac{kQq}{0.55 - x} = \frac{kQq}{0.55} + 1/2mv^2?
 
  • #11
flower76 said:
So my equation would be:

\frac{kQq}{0.55 - x} = \frac{kQq}{0.55} + 1/2mv^2?
That's correct.
 
  • #12
Doc Al said:
:eek: You might want to reconsider that, Hoot.

Wait a minute Doc, I think that I am correct. Why would it be different? The moving particle is doing work against the electrostatic force, therefore kinetic energy will be converted into potential.

~H
 
Last edited:
  • #13
So I plugged in the numbers based on my final equation and got 1.05m. So this means that the fired particle goes past the fixed charge before it would turn back, does this make sense?
 
Last edited:
  • #14
Just for reference, when using my formula I obtained 16.6cm for the distance of closest approach. I'm not sure whether it is correct though, I am waiting for Doc Al to respond.

~H
 
  • #15
flower76 said:
So I plugged in the numbers based on my final equation and got 1.05m. So this means that the fired particle goes past the fixed charge before it would turn back, does this make sense?
Of course this does not make sense. That is one reason why it is important to 'develop an intuitive feeling' for the physics and understand the numbers.

Also 1/r2 - 1/r1 does not equal 1/\Delta{r} where \Delta{r} = r2 - r1
 
  • #16
Hootenanny said:
Wait a minute Doc, I think that I am correct. Why would it be different? The moving particle is doing work against the electrostatic force, therefore kinetic energy will be converted into potential.

~H

Hoot, I don't think Doc was referring to the conservation of energy, but rather how in your earlier post you used the change in distance to evaluate the change in pot. energy and thereby the change in KE .
Compare the first part of that post ( where you wrote the proper expression ) and the second part, where you use change in distance .
They don't match, do they ?:smile:

Arun

Edit : Astronuc's faster than me
 
  • #17
Ahh, yes apologies everyone. I forgot that the particle started at a point in the electric field. I have edited by eroneous post.

~H
 
  • #18
So I am really confused now.

The equation should read:

\frac{kQq}{0.55} - \frac{kQq}{x} = E_{k}?

And Ek is \frac{kQq}{0.55} + 1/2mv^2?

Therefore \frac{kQq}{0.55} - \frac{kQq}{x} = \frac{kQq}{0.55} + 1/2mv^2 ?
 
  • #19
flower76 said:
So I am really confused now.

The equation should read:

\frac{kQq}{0.55} - \frac{kQq}{x} = E_{k}?

And Ek is \frac{kQq}{0.55} + 1/2mv^2?

Therefore \frac{kQq}{0.55} - \frac{kQq}{x} = \frac{kQq}{0.55} + 1/2mv^2 ?

No, this is wrong.
KE is always 1/2mv^2
The expression should equate so :

Initial PE + Initial KE = Final PE (no KE)
 
  • #20
flower76 said:
So I am really confused now.
Your equation that I quoted in post #11 is correct.

Hoot: As arunbg and Astronuc have pointed out, it was your mathematical step that I was objecting to. Nothing that an extra cup of coffee wouldn't cure. :wink:
 
  • #21
Hootenanny said:
You're almost there. Using conservation of energy;

U_{final} = E_{k} + U_{intial}

\frac{kQq}{x} = E_{k} + \frac{kQq}{0.55}
That last equation should be:
\frac{kQq}{0.55 - x} = E_{k} + \frac{kQq}{0.55}

x is the distance the particle moves, not the final separation of the particles.
 
  • #22
Doc Al said:
Hoot: As arunbg and Astronuc have pointed out, it was your mathematical step that I was objecting to. Nothing that an extra cup of coffee wouldn't cure. :wink:

Yeah, I've sat down and figured it out. I wasn't fully concentrating on the post (I'm actually say here typing up revision notes and I just flick over to PF when I get bored and need a break). Guess I've learned my lesson about not fully concentrating.

Again, apologies to all for the confusion. As for the cup of coffee, that sounds like a grand idea.:-p

~H
 
  • #23
Alrighty based on the equation:

\frac{kQq}{0.55 - x} = E_{k} + \frac{kQq}{0.55}

I found that x is equal to 0.32m, which is the distance that the particle moves before stopping.

Does this look a little better? This number seems more appropriate to me.
 
  • #24
Wow this thread has grown quite a bit since my last visit :smile:

I know you guys have it all figured out now, but why not plug in the values only when you must? As Astronuc said, the physical feel is most important. Anyway I just felt like whining so I did it :biggrin:
 

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