- #1
anhnha
- 181
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I have just read a thread here:https://www.physicsforums.com/showthread.php?t=382031
Here is part from the thread.
Here is part from the thread.
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?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.