Calculating Work to Push a Charge from Infinity

[anhnha]

I have just read a thread here:https://www.physicsforums.com/showthread.php?t=382031
Here is part from the thread.

Okay, here is one of those conservative, open systems, where it doesn't really specify who or what is doing the work (moving the charge in this case). So put yourself in there! Now you are the one who is going to push that charge in from infinity.

Imagine yourself pushing this positive charge toward another positive charge. You are the one putting work and effort into this. So you are giving energy to the charge which you are moving (or rather the system in general).

Let's discuss the forces. Yes, there is a force emanating from the fixed charge that is in the opposite direction of movement. But there is another force, at least as large, coming from you! Your force wins because you are able to push the charge forward. So both F and s are in the same direction. You have done positive work on the charge, and it has gained positive energy.

Switching to the case where the charges are oppositely charged, there is an attractive force between the charges. In this case, the charge is pulling you along with it. Think of a big dog on a leash that pulls you along with it. The work that you try to do pull back is futile. The system is doing work on you. In this case, your pulling force is in the opposite direction the charge ends up moving (i.e. your force loses this round). Thus the work that you do on the charge is negative. In other words, the system has done work on you, thus the system has lost potential energy.

Please see the bold sentence in the quote. In this case, does it mean that my force(push) is larger than the push force from fixed charge. If so, how can I calculate the work that I used to push a charge from infinity to a specific point?

 

[milesyoung]

In this case, does it mean that my force(push) is larger than the push force from fixed charge.

In the context of the post you quoted from, the charge has some finite velocity in the direction you're pushing it (let's assume its movement is 1-dimensional in spatial coordinates):
(1) If your applied force was greater in magnitude than the Coulomb force on the charge, it would accelerate in the direction of your applied force.
(2) If your applied force and the Coulomb force were equal in magnitude, the charge would remain in motion at a constant velocity.
(3) If your applied force was less in magnitude than the Coulomb force on the charge, it would accelerate in the opposite direction of your applied force.

(1) and (2) would let you displace the charge from its initial position to a point arbitrarily close to the charge you're moving towards.

If so, how can I calculate the work that I used to push a charge from infinity to a specific point?

In general, you'd calculate it using a path integral. Do you have a good book that treats subjects like electric fields and potential?

 

[anhnha]

Thanks, I read some books calculating the work. In all case, they say that the force that I push the charge and Coulomb force are equal as your #2 case above. Then they use path integral from A to infinity of Coulomb force.
But what I on't understand is that as in your #2 case, two forces are equal and the charge move with constant velocity. How the charge get the velocity first? Because the charge starting at rest, when it changes to a constant velocity, the two forces are no longer equal during this time and the work done calculating using the condition two forces are equal is not right.

 

[slider142]

But what I on't understand is that as in your #2 case, two forces are equal and the charge move with constant velocity. How the charge get the velocity first? Because the charge starting at rest, when it changes to a constant velocity, the two forces are no longer equal during this time and the work done calculating using the condition two forces are equal is not right.

In the post above, the postulate is that the charge starts with some finite velocity, not necessarily at rest. In an arbitrary Newtonian inertial reference frame, where Newton's laws are valid, every object with 0 net force on them has some constant velocity, which may be 0 if you choose to work in that frame. In case #2, since there is 0 net force on the charge, the charge remains at its initial velocity, which may be "at rest", if that is the frame you choose to work in.

 

[milesyoung]

Thanks, I read some books calculating the work. In all case, they say that the force that I push the charge and Coulomb force are equal as your #2 case above. Then they use path integral from A to infinity of Coulomb force.

Ah ok, I think I know the examples you're talking about. I suspect they're all purposed for deriving an expression for the potential energy of a system that has some conservative field associated with it.

In your examples, they assume that the object being displaced starts out with some velocity and doesn't accelerate, which means the applied force and the conservative force acting against it are always equal in magnitude. That way, the system has no change in kinetic energy, so the work done on the system must equal its change in potential energy (assuming no nonconservative forces act within it etc.). It's then straightforward, since they have an expression for the conservative force, to derive an expression for the potential energy of the system.

Now, that's just a way to derive an expression. It doesn't mean the applied force and the conservative force always have to be equal in magnitude and opposite in direction.

But what I on't understand is that as in your #2 case, two forces are equal and the charge move with constant velocity. How the charge get the velocity first? Because the charge starting at rest, when it changes to a constant velocity, the two forces are no longer equal during this time and the work done calculating using the condition two forces are equal is not right.

It doesn't matter that they're not equal in magnitude at some point in time. You can still calculate the work done just fine. An example:

I have a book lying on the ground - everyone likes an example with a book and gravity. I apply a constant force of 1 million Newtons in an upright direction. At an altitude of 1 km (I can fly btw.) I have done work equal to 1*10⁶*1000 J. That's a lot more than the magnitude of the work done by gravity, but I'm pretty sure the book is also approaching lightspeed.

 

[anhnha]

Thanks for helpful answer.

In your examples, they assume that the object being displaced starts out with some velocity and doesn't accelerate, which means the applied force and the conservative force acting against it are always equal in magnitude.

I feel relieved knowing that. And I have just found this (which was confusing me a lot):

"We can calculate the work done by a force on an object, but that force is not necessarily the cause of the obect's displacement. For example, if you lift an object, (negative) work is done on the object by the gravitational force, although gravity is not the cause of the object moving upward!"