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An equipotential surface is one in which all the points are at the same electric potential. If a charge is to be moved between any two points (say from point A to point B) on an equipotential surface, according to the formula $dW = q\cdot dV$, the work done becomes zero.

My question is, how to move a particle without doing any work?

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An equipotential surface is one in which all the points have the same potential. Not necessarily electric, I think. It could be, say, gravitational. –  Kaz Apr 16 '13 at 0:08

5 Answers 5

yes to move the particle you need to accelerate it which means you are applying a force and the force does some work but the point is that at the destination you should do some negative work to stop the particle so in the absence of other forces your total work is zero.

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Can you maybe add some interpunction, capitals etc? –  Bernhard Apr 15 '13 at 9:13
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Right, Richard. Moreover, it's also true that the work you have to do in this way may be made arbitrarily small by moving the body at a lower maximum speed. The maximum energy you borrow - and then repay - is given by the maximum kinetic energy the particle reaches during the transfer, $mv_{max}^2/2$, and $v_{max}$ may be chosen arbitrarily low. –  Luboš Motl Apr 15 '13 at 9:46

I think there are two sources of confusion here:

  1. How potential surfaces relate to forces.
  2. How work is defined on a potential surface

You are talking about an electric potential but all potential surfaces work in the following way:

The (negative) gradient of the potential surface is the force acting on the particle. So, in an equipotential (flat) surface, particles feel no force (the gradient is zero everywhere). No force $\implies$ no acceleration but particles will continue their trajectory if they start with some initial momentum. So particles can (and do) move without work being done.

Now there is an easier way to calculate work done if you know the start and end points of the particle trajectory on the potential surface: work done is merely the difference between the potential at the start and end points (the potential difference, or when dealing with electric fields, the voltage). This can be calculated without any knowledge of the path the particle took between these two points.

Now lets link everything together. A particle is at rest in an equipotential. You decide to move it to another point. You accelerate it, allow it to continue for a while and then decelerate it to rest. How is this compatible with the above two paragraphs?

  • By accelerating the particle, you apply a force. You can't have a force without a potential gradient so the particle is no longer in an equipotential, it is in a downward sloping potential. You do work on the particle during this stage.
  • When it stops accelerating, it is once again in an equipotential and moves at constant speed.
  • As the particle begins to be decelerated, it is again, no longer in an equipotential. This time the slope is positive. The work done initially is given back to you exactly. The particle returns to rest at a different point in space.
  • The particle sits happily in an equipotential at a different point to before. No work has been done as all energy given to the particle has been recovered.
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Just wondering if a reason could be provided for the thumbs down. Always looking to improve my answers :) –  ejrb Apr 18 '13 at 11:11

When a charge is moved from one point to another along some equipotential surface, it is true that the work done on the charge by the electric field is zero. Additional work done on the particle over this path depends on how the particle is moved. If the particle is initially moving in the correct direction, then obviously no work is required. This is why the Moon is able to maintain its motion along an (approximate) gravitational equipotential without an army of angels diligently pushing it on it's path.

But if you are moving a particle from one position at rest to a second position at rest, along the equipotential, some external agent supplying work is needed to start and stop the particle. However, you in principle you could make this motion arbitrary slow over arbitrarily long times, so that at any point along the path it's kinetic energy is indistinguishable from zero, hence zero change in kinetic energy along the path, hence zero work necessary.

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It's a matter of definition. Equipotential means same potential everywhere. The amount of work done on a charge Q when moving it from potential V 1 to V2 is (V1 - V2 )*Q. Since on an equipotential V1 is equal to V2 the net work is zero.

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If we move an object between two points where the potential is the same, it means we have not done any net work related to the particle's position in the field. It does not mean that we have not spent additional energy.

For instance, suppose the field is gravitational, and we are moving a mass in space between two points in the field that have the same potential. Suppose we use thrusters to accelerate the mass, and again to stop it as it reaches its new position. We're doing extra work to build up kinetic energy, which is quite separate from the object's potential energy. Furthermore, the energy is wasted, because even more energy is spent to stop the object. The net work against gravity is zero, but there is a net energy expenditure.

Note that an object may simply be moving already. That is to say, suppos you have a point mass in a perfectly circular orbit around another point mass moves through an equipotential surface without any work being done on that mass.

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