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I'm having some difficulty understanding this paragraph from my physics textbook regarding electrical potential energy. It is saying that the difference in electrical potential between point 'a' and point 'b', where the potential at point 'a' is greater than point 'b', can be interpreted as the amount of work needed to move a particle from point 'b' to point 'a'. I can kind of understand it intuitively, but there are some parts in the paragraph below that I don't understand. First, I don't understand by why we must "move the object slowly" and also what does it mean in the parenthesis "so as not to give it any kinetic energy". Why do we have to move it slowly, and how can you move something without giving it any kinetic energy? Wouldn't you be giving it a certain velocity if you were to move it no matter the speed at which you are doing it? Even if is constant acceleration, there would still be velocity unless it was not moving at all right? Second, it says that to do this we have to apply an external force that is "equal and opposite to the electric field force." What I don't understand about this is that if you were to apply an equal and opposite force to the charged particle, wouldn't it not move at all since the net force is zero? The textbook keeps on saying that the external force is just the negative of the electric field force, and that makes me think that there's no way the particle will move if that's the case. Maybe I'm just not reading it correctly, but to make sense of what it is saying, I was thinking that the force due to the electric field force isn't constant since as the charged particle moves from point 'a' to point 'b' due to the electric field force, the electric field force decreases due to the distance increasing between the charged particle and point 'a', as it moves to point 'b'. So if we apply an external force pushing from point 'b' to point 'a', we must start with a force in the opposite direction that is greater than the electric field force on the charged particle at point 'b' in order to get it to start moving towards point 'a', and then the external force should decrease as the charged particle approaches point 'a' so that the greater force from the electric field on the charged particle at point 'a' will bring the charged particle to a stop at point 'a'. So the force on the charged particle by the electric field force is a decreasing exponential from point 'a' to point 'b', and the force on the charged particle by the external force from point 'b' to point 'a' is also a decreasing exponential, but both will amount to the same amount of work. Is this the correct interpretation? Sorry for the super long question. Any insight is much appreciated!

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If the charged particle is accelerated to a substantial speed, it acquires also substantial kinetic energy and then total work done is not equal to change of potential energy, but to change in total energy. This includes contribution due to potential energy and due to kinetic energy. So to bring work and potential energy into simple relation, one must eliminate kinetic energy and that's why the motion has to be slow.

The forces are assumed equal because external force that is opposite in direction but equal in magnitude to the electrostatic force is enough to accomplish the process where the work gets done. Of course in real life the external force will have to be slightly stronger, so as to accomplish this work in reasonably short time. But the difference of the two forces can be arbitrarily small.

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  • $\begingroup$ Can you clarify how the total work can be a change in total energy? I thought total energy is always conserved. How can work be the result of a change in the total energy? $\endgroup$ Commented Oct 8, 2021 at 21:26
  • $\begingroup$ To my knowledge, work is defined as the change in kinetic energy or as the negative of the change in potential energy. Are you saying that if we keep the change in kinetic energy very small by keeping the motion small, the change in kinetic energy will technically be zero, but the change in potential energy remains and thus work is only defined as a change in potential energy? Meaning that moving the charged particle from point 'b' to point 'a', the work done is only a change in potential energy? Thanks again $\endgroup$ Commented Oct 8, 2021 at 21:32
  • $\begingroup$ By total energy I mean potential energy plus kinetic energy of the particle. It is not conserved when external force does work on the particle. Work is defined is force times distance. It also equals numerically to change in potential energy, if kinetic energy is kept without change. $\endgroup$ Commented Oct 9, 2021 at 10:05
  • $\begingroup$ Ah I see. I'll look up how non conservative forces affects energy of a system. Thank you! $\endgroup$ Commented Oct 9, 2021 at 19:46
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    $\begingroup$ @JerryHsieh yes. $\endgroup$ Commented Oct 11, 2021 at 13:54

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