1mgx - ∫E(x) dx =1/2 mv2
v = vel at equilibrium
now what ?
Suppose a disc of radius a placed on the ground with uniform surface charge density s with its axis vertical.A charged particle is dropped from a height H from the ground along the axis of the disc.Initially the particle is at rest so initial kinetic energy is equal to zero.Then see the point when the electric force on the particle is equal to the gravitational force.At that point velocity of the particle is zero.Hence the change in kinetic energy is equal to zero.So the work done by the net external force should be zero.But the resultant force is causing a net displacement in the downward direction.So there has to be some work done by the net external force.Please explain this.
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67 Answers
1Yes you have got my point.Do not take the time after equilibriumJust when from the particle is dropped and then to the equilibrium.And then apply F.dx and WET.
1why are taking some v as at equilibrium you said that velocity is equal to zero.
1?????why velocity at equilibrium = 0 ? when did i say that ?
net force = 0 implies change in momentum is zero , it doesnt imply momentum itself is zero
1Yes it does not imply that momentum itself is zero.But how can you say that the is momentum is not zero at equilibrium?
1Two equal and opposite forces are acting so the particle will not move.So the momentum is zero.Have you understood?
1see this is analogous to shm .... in that , at equilibrium vel = max !!