Finding the Distance: Charge Particle Motion

[flower76]

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?

 

[pizzasky]

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.

 

[maverick280857]

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.

 

[flower76]

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?

 

[Doc Al]

It's almost right. What's "r"?

 

[flower76]

r is the initial distance between the charges = 0.55m

 

[Hootenanny]

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

 

[Doc Al]

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?

 

[Doc Al]

However, it would be easier to consider the change in distance;

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

Since $\Delta r = 0.55 - x$ ;

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

You might want to reconsider that, Hoot.

 

[flower76]

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:

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

 

[Doc Al]

So my equation would be:

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

That's correct.

 

[Hootenanny]

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

 

[flower76]

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?

 

[Hootenanny]

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

 

[Astronuc]

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

 

[arunbg]

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 ?

Arun

Edit : Astronuc's faster than me

 

[Hootenanny]

Ahh, yes apologies everyone. I forgot that the particle started at a point in the electric field. I have edited by eroneous post.

~H

 

[flower76]

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

 

[arunbg]

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)

 

[Doc Al]

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.

 

[Doc Al]

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.

 

[Hootenanny]

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.

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.:tongue2:

~H

 

[flower76]

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.

 

[maverick280857]

Wow this thread has grown quite a bit since my last visit

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