A unit positive charge $q$ is placed in an electric field caused due to a positive charge $Q$. $q$ experiences force of repulsion $F_R$. We apply an external force (FEXT) opposite to $F_R$. The test charge moves in the direction opposite to electric field. Work is done opposite to the field. This work done is stored as potential difference.

Now my doubt is below:

We know that $F_R=-F_\mathrm{ext}$, so two equal and opposite forces are acting on the charge $q$. If they are equal and opposite, they should cancel out each other know? When how is work done against the field? Why is the charge $q$ moving in the direction of external force if the repulsive force (equal in magnitude) is opposing it?

  • $\begingroup$ If the test charge is moving, as you say, how can you conclude that $F_R=-F_{\text{ext}}$? Actually, $F_R>-F_{\text{ext}}$, and so work done by the external force is positive. The work done by the repulsive force is negative, which gets stored as potential energy. $\endgroup$ – FreezingFire Apr 25 '16 at 14:37
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    $\begingroup$ @FreezingFire if the charge is moving at constant speed, then the magnitude of the forces will be equal. $\endgroup$ – garyp Apr 25 '16 at 14:41
  • $\begingroup$ @garyp That's what happens when i hurry. Of course you are right! $\endgroup$ – FreezingFire Apr 25 '16 at 14:44
  • $\begingroup$ @FreezingFire I've been there many times myself. :) $\endgroup$ – garyp Apr 25 '16 at 14:58
  • $\begingroup$ @FreezingFire. Charge can considered to be in quasi-equilibrium. $\endgroup$ – Prayas Agrawal Apr 25 '16 at 17:02

Potential energy is associated with a system, not a particle. One must calculate potential energy for pairs of interacting particles, not individual particles.

$F_{ext}$ is a force external to the system. It can do external work and raise the energy of the system. $F_R$ is work internal to the system, and does not increase the energy of the system.

But there is a connection which can make the ideas confusing. The definition of potential energy is $\Delta PE = -W_\mathrm{ext}$. But sometimes, as in your example, as a consequence of Newton's Third Law the internal work can be calculated by examining the external forces.

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