Positive work:direction of external force

[gracy]

If an electron is brought near the proton from infinity,potential energy of the electron decreases so work done must be positive because the change in potential energy is the NEGATIVE of the work done .But how ?I mean for work done to be positive, force and displacement should be parallel to each other here

so is the external force which is responsible for doing the work is in the same direction as of direction of displacement?

 

[andrewkirk]

I presume you are not talking about a quantum analysis of this, as the concept of an electron approaching a photon assumes an exact measurement of position.

So I imagine you're just talking about a classical negatively charged particle approaching a classical positively charged particle. In that case the force is attractive (opposites attract) so the force on the -ve particle is in the same direction as the displacement - both being towards the +ve particle.

That gives positive work.

 

[gracy]

so is the external force which is responsible for doing the work is in the same direction as of direction of displacement?

I am not talking about attractive force rather I am referring to the external force which is responsible for doing the work

 

[andrewkirk]

There is no need for an external force. The -ve particle will fall towards the +ve one without an external force, and positive work will be done in the process. The work will emerge as kinetic energy.

If you want to introduce an external force you need to tell us what it is doing. Is it for instance pushing the -ve particle away from the +ve one in order to slow its descent? In that case you can no longer use a principle that ' the change in potential energy is the NEGATIVE of the work done' unless you also take into account the change in potential energy of the system providing the external force. So you have two options:

1. No external force. The PE net decreases by w. The work done by the attractive force is +w, and the KE of the particle increases by w.

2. An external force is pushing the -ve particle away. Then the external force does work -w, so the net work done is
+w (by the attractive force) + -w (by the external force) = 0
This matches the change in KE of the particle, which is zero.
The PE of the particle decreases by w, which is exactly offset by the increase in PE of the system providing the repulsive force (assuming no work is dissipated as heat). So the change in PE is zero, which matches the work done.