External Forces and Potential Difference

[Abu]

Homework Statement

The work done by an external force to move a -8.0 uC charge from point a to point b is 25*10^-4 Joules. If the charge was started from rest and had 5.2 * 10^-4 Joules of kinetic energy when it reached point b, what must be the potential difference between a and b?

Homework Equations

ΔU= Δv *q

The Attempt at a Solution

My attempt was basically trying to understand what the question is asking from me:

From what I understand from this problem statement, there is an electric field force that is pushing the negative 8 uC charge away from point B, so as a result the charge requires work from an external force to get to point B.

The external force is pushing the negative charge against the force of the electric field, and once it reaches point B, it does not stop and hold the charge there. I am assuming the charge is still being pushed, so that means there is kinetic energy.

My question is if the 25*10^-4 Joules of work done by the external force is the total work done, and if so, how come some of it is transferred to kinetic energy - is it because the charge is still moving so that means not all of it can become electric potential energy, and it is only when the charge is held still when all of the energy is electric potential energy?

Thank you very much.

 

[haruspex]

My question is if the 25*10^-4 Joules of work done by the external force is the total work done

Yes.

is it because the charge is still moving so that means not all of it can become electric potential energy, and it is only when the charge is held still when all of the energy is electric potential energy

Not really.
The work that has gone into the potential energy depends only on the field and the change in position. It has no dependence on the speed. The point is that the external force did some total of work, some of which went into the potential energy and some into the change in kinetic energy. Stopping the movement would not magically put that KE into potential energy. What you could do is allow the acquired KE to be spent moving further against the field, with no continuation of the external force.

 

[Abu]

Yes.

Not really.
The work that has gone into the potential energy depends only on the field and the change in position. It has no dependence on the speed. The point is that the external force did some total of work, some of which went into the potential energy and some into the change in kinetic energy. Stopping the movement would not magically put that KE into potential energy. What you could do is allow the acquired KE to be spent moving further against the field, with no continuation of the external force.

Thank you for the clarification. Let me just see if I understand fully now... So the charge moves against an electric force pushing it in the opposite direction because of the external force provided. Because of the external force, some of the work is put into potential energy (because it is getting closer to the source of the electric force) and some of that work is put into kinetic energy (because now the charge is moving). If the external force were to stop being given, the kinetic energy it was given during its motion will slowly be converted to potential energy as it slows down. Do I understand it now? And am I correct in the assumption that the charge is moving against a force that is repelling it?

Thank you.

 

[haruspex]

charge moves against an electric force pushing it in the opposite direction because of the external force provided. Because of the external force, some of the work is put into potential energy (because it is getting closer to the source of the electric force) and some of that work is put into kinetic energy (because now the charge is moving).

Yes.

If the external force were to stop being given, the kinetic energy it was given during its motion will slowly be converted to potential energy as it slows down.

Not necessarily. The force need not be constant, and the field could be non-uniform. Consequently, we do not know whether the momentum at the end is taking it to a higher potential energy or a lower.
E.g. imagine the field being generated by a point charge, and a constant force with a line of action that misses the point charge to one side. At first, force is doing work against the field, but eventually it will be going the other way. Turning off the force will leave the particle gaining speed as it accelerates away from the charge.

 

[Abu]

Yes.

Not necessarily. The force need not be constant, and the field could be non-uniform. Consequently, we do not know whether the momentum at the end is taking it to a higher potential energy or a lower.
E.g. imagine the field being generated by a point charge, and a constant force with a line of action that misses the point charge to one side. At first, force is doing work against the field, but eventually it will be going the other way. Turning off the force will leave the particle gaining speed as it accelerates away from the charge.

Like this, right?

So you are saying that initially the charge is being pushed directly against the force from the electric field, but once the charge gets to a point below the point charge emitting the electric field, it will immediately be accelerated downwards and not necessarily slow down first.

It can be assumed that indeed the electric field force is not constant because as you get closer to the point charge the repulsive force will be greater, and the external force is not constant too because more external force will have to be provided to overcome the increasing repulsive force... But for the sake of introductory physics we can assume that the charge is moving directly head on to the other charge, in which case what we said about the charge slowing down then being emitted west will then apply, right?

 

[haruspex]

So you are saying that initially the charge is being pushed directly against the force from the electric field, but once the charge gets to a point below the point charge emitting the electric field, it will immediately be accelerated downwards and not necessarily slow down first.

well, not directly against even at the start, so at the same time that the force is pushing it one way the fixed charge is pushing in another direction. Consequently the trajectory will curve around.
If the force is weak compared to the electric force, it may be that the potential energy never increases, but decreases right from the start. Nevertheless, it remains true that the work done by the external force is ΔKE+ΔPE.

 

[Abu]

well, not directly against even at the start, so at the same time that the force is pushing it one way the fixed charge is pushing in another direction. Consequently the trajectory will curve around.
If the force is weak compared to the electric force, it may be that the potential energy never increases, but decreases right from the start. Nevertheless, it remains true that the work done by the external force is ΔKE+ΔPE.

Thank you very much for your time and patience. You are great at explaining physics. Thank you

Are we able to just quickly address the last part of my statement, starting from the "sake of introductory" part to the end?

 

[haruspex]

for the sake of introductory physics we can assume that the charge is moving directly head on to the other charge, in which case what we said about the charge slowing down then being emitted west will then apply, right?

Yes, but of course no finite force would be enough to push the particle all the way to the fixed charge.

 

[Abu]

Yes, but of course no finite force would be enough to push the particle all the way to the fixed charge.

Thank you, sir.