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In the textbook it says that no work is required to move a charge from one point to another on an equipotential surface. Do they mean work by the electric field or work by anything? Because clearly the object cant just magically move sideways with nothing.

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"it takes no work" in the same sense it takes no work to move an object on a perfectly frictionless, flat surface. It is true and theory, but moving an object requires accelerating it at least a little bit, which requires some work, as you point out.

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  • $\begingroup$ Agreed. will remove my comment in the meanwhile. $\endgroup$ – ZeroTheHero Sep 4 '17 at 1:24
  • $\begingroup$ The work done in the non ideal case is not against the electric field (unless your path is not exactly along the equipotential). So, if you want to narrow your focus to only electric forces, then there is no work done in that category, which is the essential point. $\endgroup$ – EL_DON Sep 4 '17 at 2:32
  • $\begingroup$ P.S. the object could already be in motion, with continued motion along the equipotential requiring no work be done. $\endgroup$ – EL_DON Sep 4 '17 at 16:09
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It takes work to transfer kinetic energy into the charged object and get it moving, sure. But if the object was already moving, it wouldn't lose any energy by moving along the equipotential. Furthermore, you could decrease the work needed to stop and start by just moving slower, with no work needed in the limit of infinite time taken. The work of starting and of stopping can cancel each other out so that all your energy can be recovered in an ideal system. So if work is needed to move a charge along an equipotential, it isn't because of the electric field.

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  • $\begingroup$ "So if work is needed to move a charge along an equipotential, it isn't because of the electric field." So its due to something else? Is my thinking correct?: If i drop a positive charge near an electric field, it should go radially outwards. If i wanted to change its position on an equipotential line (sometime when the charge is getting repelled) then i would need to apply a force directly perpendicular to its current path (and thus do work) and during the whole process of moving the charge sideways (to another position in an equipotental line) the electric field does no work $\endgroup$ – LTM Sep 4 '17 at 2:04
  • $\begingroup$ You could do work against friction or gravity or something while moving along the equipotenyial of the electric field, yet no work is done against the electric field. $\endgroup$ – EL_DON Sep 4 '17 at 2:30
  • $\begingroup$ Ok, but to move along the equipotential line in the first place would require some sort of force by something (and as a result work) correct? Since, in order to move along an equipotential, you would have to move sideways. $\endgroup$ – LTM Sep 4 '17 at 2:46
  • $\begingroup$ Yes, but that work isn't against the field and you can reduce it to being arbitrarily small business going very slowly. You can also fully recover the work done when you come to a stop. The sideways motion could also just be continuous, with no stop or start, and go on forever since the electric field isn't slowing it down. $\endgroup$ – EL_DON Sep 4 '17 at 3:41
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If the particle is on an equipotential surface, then that means there is no force from the electrostatic field on your charge, while moving along that surface. If there are no external forces, that means that locally momentum in that direction is conserved, just consider Newton's laws: $$ \frac{\text{d}\boldsymbol{p}}{\text{d}t} = \boldsymbol{F} = 0 $$

However in moving your particle, you would change the velocity, so we require $$\Delta \boldsymbol{p} = m_q \Delta \boldsymbol{v} \neq 0$$ Hence, to move your particle along an equipotential surface, you don't need to do any work, but you do need supply a change to the particle's momentum. Hence the particle can't spontaneously change its trajectory while on an equipotential surface. The issue here is when changing the momentum, in general you will change the energy too, and end up doing some work with respect to whatever external field you're using to push or pull your charge around.

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  • $\begingroup$ Acceleration obviously DOES require work. $\endgroup$ – EL_DON Sep 4 '17 at 2:21
  • $\begingroup$ @EL_DON Only if you change $|\textbf{v}|$. Which is why I said in general when changing the momentum, you will also give the particle some energy. $\endgroup$ – CDCM Sep 4 '17 at 11:07
  • $\begingroup$ Yes, that's true, but acceleration perpendicular to velocity (circular motion) is not the case implied by your wording. You are talking about the case where you push the particle along the equipotential and then push or pull it to a stop. This changes that magnitude of momentum and explicitly requires work. Arguments about changing only the direction of momentum are not applicable here. $\endgroup$ – EL_DON Sep 4 '17 at 15:14
  • $\begingroup$ @EL_DON That's why I say in general it will require work to be done. But there is the possibility of it moving, without work, and what stops that from spontaneously happening, is conservation of momentum, so I think that is pretty important to discuss. I'm not talking about a stationary charge either. Any charge moving in a field, based on solely energy alone should be freely able to move along an equipotential line, mid-trajectory. However this would break local conservation of momentum, so is not observed. $\endgroup$ – CDCM Sep 4 '17 at 16:00
  • $\begingroup$ Kinetic energy = $p^2/2m$, which covers momentum just fine without separate discussion in this case. We are not talking about a closed system, so reference to momentum conservation is not useful to understanding the problem. Your answer still contains a false statement, in that you discuss changing momentum without doing work and the case is clearly not something like circular motion where only the direction, but not magnitude, of momentum changes. Even for circular motion, you still have to get started of not already in motion, requiring work be done. $\endgroup$ – EL_DON Sep 4 '17 at 16:06

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