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if work is done along the direction of force,then the work is regarded as positive work and if work is done in a direction opposite to the direction of force then it's regarded as negative work.Whereas in electrostatics, if a positive charge is brought near a positive charge(which produces an opposing force) the work(electric potential) is regarded as positive and if the same positive charge is brought near a negative charge then the its regarded as negative work(potential).Just the opposite

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  • $\begingroup$ I think you are misinterpreting the sign conventions for work done by system and work done by external force...Try this answer by steeven here physics.stackexchange.com/questions/248629/… He has explained very clearly and I have added some doubts too in comments $\endgroup$ – user184271 Mar 2 '18 at 12:08
  • $\begingroup$ The sign of the potential doesn't really mean anything. You can add an arbitrary constant to the potential and it doesn't affect the behavior of the system, so you can make whatever part you want positive or negative by adding the right constant. The part of the potential that has physical meaning is the slope, which corresponds to the force on the particle in that potential. $\endgroup$ – probably_someone Aug 23 '18 at 3:04
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Whereas in electrostatics, if a positive charge is brought near a positive charge(which produces an opposing force) the work(electric potential) is regarded as positive.

Here, the work done by the external force (which is you pushing the charge) is positive, cause you displace the charge along the direction in which you apply force. On the contrary, the work done by electrostatic repulsive force is negative ...

and if the same positive charge is brought near a negative charge then the os regarded as negative work(potential)

Here, you the force is attractive. When you are included (an external force), it always implies an opposite force to the electrostatic force. So the positive charge eventually displaces into the direction of negative charge. But you try to pull it upwards ... You need not push it towards the negative source as the force here is already attractive ...

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  • $\begingroup$ One thing the op might find helpful is that conservative forces try to decrease energy so work has to be done against that force for increasing potential energy so hence the positive sign $\endgroup$ – user184271 Mar 2 '18 at 12:14
  • $\begingroup$ @harambe...Why do conservative forces decrease energy...? Is it because a stable system has least potential energy ...? $\endgroup$ – Nehal Samee Mar 2 '18 at 12:39
  • $\begingroup$ I am just a school going student so I just go by my simple definition that is negative and positive work for conservative forces increase and decrease energy of the system respectively.Also any object if left free would always try to go from higher potential ti lower potential which is what conservative forces do but I am not that sure.You can check here physics.stackexchange.com/questions/249430/… $\endgroup$ – user184271 Mar 2 '18 at 13:12

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