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Assume I have a test charge of + 1 coulomb.

  • If I move the test charge from a point with potential of +9 volt to another point with potential of -1 volt then the work done by me is -10 joule. This negative numbers means I actually do not need to do such a work because the test charge can move automatically.

  • If I move the test charge from a point with potential of -9 volt to another point with potential of +1 volt then the work done by me is +10 joule. This positive numbers means I actually do need to do such a work because the test charge cannot move automatically.

Question

  • If I move the test charge from a point with potential of $v$ volt to another point with the same potential of $v$ volt then the work done by me is 0 joule. What does this zero mean physically?
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  • $\begingroup$ It means that the change in the kinetic energy of the charge is zero. $\endgroup$ – Farcher Sep 4 at 7:40
  • $\begingroup$ @Farcher: I assume for all cases I mentioned above, the change in kinetic energy is kept zero. $\endgroup$ – Money Oriented Programmer Sep 4 at 7:42
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    $\begingroup$ If "you" are moving a charge along an equipotential at constant speed, you don't have to exert a force along the direction of motion (i.e. along the equipotential) to keep the charge moving, just like a marble rolling along a horizontal track would continue to roll (absent friction)). However, you will be exerting forces perpendicular to the equipotential in order to both counteract the electric forces on the charged particle that also act perpendicular to the equipotential and to "counteract the inertia" of the charged particle that makes it want to move in a straight line. $\endgroup$ – march Sep 4 at 16:05
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Let's take the Earth's orbit around the Sun as an example, and assume it's a perfect circle. In this case there is no work done as the Earth goes round, i.e. it can do so forever without requiring (or expending) energy.

The same applies to your scenario, and for all cases where work done is zero.

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