Determining the final position of test charge with multiple potentials

[bluedevil09]

Homework Statement

Consider the following arrangement of potentials and a test charge (not pictured) located at infinity. - To which location will a test charge of +3 microColumbs travel?

The potentials are as follows:

• (-1V) top left
• (-7V) top right
• (+1V) lower left
• (+4v) lower right

They do not line up(top left is further left than bottom left, etc.), and no distances are given

Homework Equations

force = q * E (charge * field strength)
All of the other equations I can think of require distance

The Attempt at a Solution

I honestly don't know what to do here. If I was given distances I would be able to calculate the force on the charge, but since I am not I do not know how to calculate where exactly it will end up. I assume it will be near the -7V charge, but that is a very general answer. If someone could shed a little light on how I could figure this out I'd greatly appreciate it
Thanks!

 

[LowlyPion]

Welcome to PF.

Isn't it a matter of which Voltage node the charge will be most attracted to if left to follow the field lines on its own?

 

[bluedevil09]

Thanks, it looks like a great place!

I was thinking that at first, but I noticed that it is a 4 point question, so I was guessing that there had to be something more to it. I don't know what else you could do with the info given though. If that was what they were looking for, it would be the -7V charge, correct?

 

[LowlyPion]

Looks like it to me too.

A +3 would mostly likely want to go steady with the -7.

 

[bluedevil09]

The next part of the question asks about the kinetic energy when it arrives. I would only have to use the -7V potential to calculate that and not any of the others, correct?

 

[LowlyPion]

The next part of the question asks about the kinetic energy when it arrives. I would only have to use the -7V potential to calculate that and not any of the others, correct?

From ∞ the net voltage is what determines the work to get there. Hence the sum of the voltages. I think the energy when it arrives will be influenced by the net of the voltages.

 

[bluedevil09]

Sorry, one last question. It also wants me to find the location with a -3 microColumb test charge. Would this travel to the +4V potential even though the net charge is negative? I have to calculate its potential energy for the next part, but I should be ok for that as long as I know where it is going. I apologize for all of the questions but I want to understand this and it is not covered at all in our lessons.

 

[LowlyPion]

If the test charge is (-), and the net voltages are (-), what incentive does it have to leave ∞?

If it doesn't leave ∞, what change in potential energy is there from 0 that would be possible.

(If you say it gains - potential then I will wonder if your charge is really at ∞.)

 

[bluedevil09]

I figured it would just remain at infinity also, but the next part gives the mass of the new test charge and asks for its speed when it reaches its final destination. If both the charge and net potential acting on it are negative and the charge is already at infinity, how can you calculate its speed? I'll type everything out for you and see what you think:

A. To which location will a test charge of +3 microColumbs travel
B. What will be its kinetic energy when it arrives
C. Find the capacitance of a capacitor that would store the same amount of energy as you found in part B
D. suppose the +3 microColumb charge is replaced with a test charge of -3 microColumbs. To which location will it travel?
E. If the new test charge has a mass of 27.3E-21 kg, what will be its speed when it arrives.

My only thought is that maybe they are talking about switching the test charge after it has been pulled in from infinity to the -7V potential. What do you thinK?

I'll note quickly that this is an AP Physics online course and some of the homework problems have proven to be literally impossible...it can be frustrating.

 

[LowlyPion]

When it arrives - from ∞ - its potential energy would be converted to kinetic wouldn't it?

ΔPE = ½mv²

But if they are talking about the -3 test charge I think ΔPE = 0.

 

[bluedevil09]

When it arrives - from ∞ - its potential energy would be converted to kinetic wouldn't it?

ΔPE = ½mv²

Right. The thing is, the net potential is -3V. The charge is now -3uC. There is no reason that it would leave infinity, correct?

 

[LowlyPion]

Right. The thing is, the net potential is -3V. The charge is now -3uC. There is no reason that it would leave infinity, correct?

Yes, I just added that thought to the last post.

 

[bluedevil09]

ah...got it. Well, I'm just going to assume that it is starting from where the +3uC charge was and going to infinity from there since I don't know what else I can do. Thanks for all of your help! I'm not used to asking questions, but I was lost on this one.

 

[LowlyPion]

ah...got it. Well, I'm just going to assume that it is starting from where the +3uC charge was and going to infinity from there since I don't know what else I can do. Thanks for all of your help! I'm not used to asking questions, but I was lost on this one.

You might want to check again to be sure the signs of the problem are correct as stated. But I think with the current statement, if the -3 is at ∞ it stays there if the net of the voltages is (-).

 

[bluedevil09]

I just double checked, the signs are correct. Since it gives me a mass and information, I'm sure that it wants a speed calculated, which means that the -3 has to move(it asks for the speed of the new test charge). From that, I'm just guessing that when they talk about changing charges in D they are replacing the +3 with -3 after it has come all the way into the potentials. The -3 would then be repelled out to infinity. Think that could be right?

 

[LowlyPion]

No. If they swap the charges, that means put a -3 at the -7 voltage, then it would arrive at the +4 voltage node having experienced an acceleration of 7v + 4v or 11 v.

This charge if it were located at the -7 would go to the +4 directly, no visit to ∞ involved. As I originally read what you posted I thought it meant start at ∞ again.

Edit: As before ΔKE = q*ΔV = ½mv²

 

[bluedevil09]

Of course...I forgot about that. What do you think I should use for calculating the speed then? I know it won't have a speed of 0(they wouldn't make it a 4pt question or give mass), so it has to be moving somewhere.

 

[JumpinJohny]

B. What will be its kinetic energy when it arrives

I'm having trouble figuring out how you solved for this problem. I understand that the kinetic energy would be equal to the potential energy, but I'm not sure how to solve for the potential energy

 

[LowlyPion]

I'm having trouble figuring out how you solved for this problem. I understand that the kinetic energy would be equal to the potential energy, but I'm not sure how to solve for the potential energy

ΔPE = W = q*ΔV

 

[JumpinJohny]

ΔPE = W = q*ΔV

But the velocity is not given, is it?

 

[JumpinJohny]

Ahhhh voltage my bad. So would the change in voltage be 3-(-7) then? So, 10?

 

[LowlyPion]

No the ΔV is -7 to +4 = 11v.

The charge is 3.

 

[JumpinJohny]

So it'd be 3*11=33 J then?

 

[LowlyPion]

So it'd be 3*11=33 J then?

That would be the work to move a +3C charge.

The test charge was +3μC I believe.

 

[JumpinJohny]

E. If the new test charge has a mass of 27.3E-21 kg, what will be its speed when it arrives.

You two never figured this one out. I'm assuming we could use the same concept for part B, except this time we'll solve for v on the kinetic energy side?

 

[LowlyPion]

You two never figured this one out. I'm assuming we could use the same concept for part B, except this time we'll solve for v on the kinetic energy side?

Well I figured it out. I just didn't post it.

And yes the work becomes kinetic energy is the method.

 

[jmwainright]

I would like to add that i am also in the same AP Physics class stated, and i would also like to express my frustration with the class. there is nothing like this in the homework or notes, and it is really hard to take these tests. Thanks to all who helped solve this problem, it helped me out emensley!