Work to move electron/proton between the plates of a charged capacitor?

[JustAStudent]

Homework Statement

A charged capacitor consists of two large flat plates, one with positive charge +Q and the other negative charge -Q. An external agent trans an electron from the positive plate to the negative and also transfers a proton from the negative plate to the positive plate. There is a constant voltage difference delta V between the plates and the transfer of a a single electron or proton does not alter that voltage.
Consider the following statements:
i.) the work done by the agent to move the electron is positive
ii.) the work done by the agent to move the proton is positive
iii.) the magnitude of the work to transfer the proton is the same as the magnitude of the work to transfer the electron.
iv.) the magnitude of the work done to transfer the electron is less than the magnitude of the work to transfer the proton.

Which of the statements are true? (MC)
a.) all
b.) none
c.) only iii
d.) ii and iv
e.) i, ii, and iii

Homework Equations

W=F*d
F=ma
F=kq/r^2

The Attempt at a Solution

I understand that i and ii are correct because they are both moving in the same direction as the force that's being applied (the external force). I don't understand is why the answer is E--because I don't understand why the forces are the same (making the works the same magnitude). If the masses of the proton and electron are different, wouldn't that make the forces different?

 

[haruspex]

the masses of the proton and electron are different,

Why would that make the forces different?

 

[JustAStudent]

Because of F=ma. If the masses are different then either the force is different or the acceleration has to compensate to make the forces be the same and I don't know why the accelerations are different.

 

[haruspex]

I don't know why the accelerations are different.

You have not provided a reason that the forces should be different, nor a reason why the accelerations should be the same.
What determines the force on a charged particle in an electric field?

 

[JustAStudent]

Okay rather than stating to me what I need to know, why don't you explain to me why I don't know what I don't know. I don't know why the accelerations are different or the same. Or if that's even relevant to the question at all. I've worked on this problem for about an hour and I don't understand it.

All I know is that the mass of a proton and an electron are different. The force on a charged particle in an electric is determined by the electric field. That's all I got.

 

[JustAStudent]

I know that the forces are the same because the work is the same. W=F*d. And I know the answer is E to this question. So I know that forces are the same I just don't understand WHY.

I also don't know if this question is asking about the work done by the e-field or the work done by the external force at this point. That's how confused I am.

 

[haruspex]

The force on a charged particle in an electric is determined by the electric field.

Good, but what else? That would have been appropriate as one of the "relevant equations".
The answer to your question can be had by combining that with this equation:

W=F*d

Note that neither one mentions mass.

the work done by the e-field or the work done by the external force

They are equal and opposite. You can suppose the particles start and end at rest, so no net work is done on them.

 

[JustAStudent]

So why is it equal and opposite? If they were equal, how could the object be moving?

 

[JustAStudent]

Good, but what else? That would have been appropriate as one of the "relevant equations".
The answer to your question can be had by combining that with this equation:

Note that neither one mentions mass.They are equal and opposite. You can suppose the particles start and end at rest, so no net work is done on them.

Good, but what else? That would have been appropriate as one of the "relevant equations".
The answer to your question can be had by combining that with this equation:

Note that neither one mentions mass.They are equal and opposite. You can suppose the particles start and end at rest, so no net work is done on them.

Also, I thought because force has mass in it's equation (F=ma) that it would be related.

 

[haruspex]

So why is it equal and opposite? If they were equal, how could the object be moving?

The net work can be equal and opposite. At the start, some acceleration is needed, though it can be as small as you like. During this phase the applied force must slightly exceed the opposing force from the field. At the end, you can remove the force and let the KE of the particle carry it the rest of the way, arriving at zero speed.

 

[haruspex]

force has mass in its equation

The F=ma equation relates force, mass and acceleration. Other equations relate force to other things, mass to other things, etc.
You could analyse the question in terms of accelerations, velocities, etc., as I described in post #10, but you do not need to. You just need the equation relating force to distance and work done.

 

[JustAStudent]

At the end, you can remove the force and let the KE of the particle carry it the rest of the way, arriving at zero speed.

Okay so in that case the total acceleration then is constant. Or 0. . . yes?

 

[haruspex]

Okay so in that case the total acceleration then is constant. Or 0. . . yes?

Yes, you can take it as zero.

 

[JustAStudent]

Yes, you can take it as zero.

So would that make the force=0 then? For both.

 

[haruspex]

So would that make the force=0 then? For both.

No, acceleration is the result of the net force. Two forces act in opposition.

 

[JustAStudent]

No, acceleration is the result of the net force. Two forces act in opposition.

I feel like I'm going in circle then at this point with Newton's 2nd law. If acceleration is 0, would that not make force 0?

I understand how you were saying that the original external force would have to be slightly greater than the electrical force for the proton/electron to move to begin with and then eventually the acceleration decelerates and the velocity is 0. If acceleration is equal to vf-vi/t and it both starts at 0 and ends at 0 would that not make acceleration 0?

Again, you said that you just need the equation relating work and force and distance (W=F*d). From there, I know that d is the same. And you said that the electrical force is equal and opposite to the external force. Which means the forces for both or the same. But then I replied saying that it means it can't move. And then you said it must start with some higher force that's greater than the e-field force. And eventually it will stop and v=0 at whatever that distance is. So, by definition, does that not mean that the net acceleration was . . . 0?

 

[haruspex]

does that not mean that the net acceleration was . . . 0?

Yes. Does that worry you? If an object starts and finishes at rest, no matter what happened in between, its average acceleration was zero.

 

[JustAStudent]

Yes. Does that worry you? If an object starts and finishes at rest, no matter what happened in between, its average acceleration was zero.

I don't think 'worry' is the right answer. It's the fact that Force=0 that confuses me if we're relating it back to Newton's 2nd law.

 

[haruspex]

I don't think 'worry' is the right answer. It's the fact that Force=0 that confuses me if we're relating it back to Newton's 2nd law.

A) that's the net force, i.e. what you get when you cancel out the applied force with the opposing electrostatic force.
B) that's the average net force. There is no reason why it cannot be positive, in the direction of the applied force, to begin with, but negative later.

 

[JustAStudent]

A) that's the net force, i.e. what you get when you cancel out the applied force with the opposing electrostatic force.
B) that's the average net force. There is no reason why it cannot be positive, in the direction of the applied force, to begin with, but negative later.

Right. I guess in order for me to answer this question, I just don't see that why I would need to think about it so in depthly in order to answer this problem. If I inherently thought the forces were equal and opposite then I would just assume that the proton or electron never moved at all.

I get that would mean that the work is negative and then it's positive and therefore the net force and the net work is 0, but conceptually that doesn't entirely make sense to me.

 

[haruspex]

the forces were equal and opposite then I would just assume that the proton or electron never moved at all.

You are still not getting the crucial difference between the net force being constantly zero and the net force being zero on average. If you apply a net force of 1N to a mass, initially at rest, for time t, then apply 1N in the opposite direction for time t then the object will come to rest; but it will not be where it started.

 

[JustAStudent]

You are still not getting the crucial difference between the net force being constantly zero and the net force being zero on average. If you apply a net force of 1N to a mass, initially at rest, for time t, then apply 1N in the opposite direction for time t then the object will come to rest; but it will not be where it started.

Okay, yes, I get that. But you weren't explaining it like that earlier. Usually when I think of Net force I think of all the forces being added up (in this case, the external and the electrical field force).

 

[haruspex]

Usually when I think of Net force I think of all the forces being added up (in this case, the external and the electrical field force).

And that is how I am using it here.

The difficulty with questions like these ( and I assume this relates to your difficulty) is the issue of KE. In this case there are two ways in which we can show it can be ignored:
- let the applied force vary, as I described, so that the particle just arrives with no residual KE
- make the excess of the applied force over the electrostatic force extremely small. The particle will still get there, with neglible speed, just taking a long time.
Either way, the work done is only that required to move the charge across the given potential.