At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
I don't understand your integrating stuff.. but.. there is no work done cause.. there is no other charge .. hence there is not other field.. and hence there is not resistance to this charge that you are bringing into this finite world.. !!.. and hence we say we didn't need to do any work.. !
and strictly speaking.. you are not CREATING field as such.. its already there.. associated with the charge.. its an intrinsic property of the charge..
why ain't anyone replying? :D was my reply so awesome? :D that it rendered you guys speechless? :D
well i guess everyone will at first think of this answer only...it most probably is right...still something fishy is going on with the field storing the energy thing.any other opinion about this,anyone???
@Mashy is right. And i don't understand why you think some energy is stored in space just because the charge is moved through it. As far as I'm concerned, moving a charge from point A to point B infinitesimally slowly is equivalent to not moving the charge at all.
You're not creating the field, it's already there. All you're doing is changing its strength in two different places (infinity and right here). Whatever work you do increasing the strength of the field here is exactly matched by the work released when the field decreases at infinity, because the interactions of the charge are identical in both places. Infinity isn't any stranger place than 5 m away, in this case, and you don't think you would do work moving the charge 5 m to the left in empty space, do you?
You certainly must do work to *create* the field in the first place, and if you are talking about *creating* a charge out of nowhere, then that will indeed require energy. If you are asking about the additional energy required to create an electric charge, beyond that required just to create an uncharged particle of the same mass, that is a subtle question, and I'm not 100% sure of the answer, if an answer is even possible.
this is a nice question.......we say there is no work done because we bring the charge under the absence of any external electric field but your point is that when u move a charge in space it creates an electric field of itself....well yes it does but doesnt get affected by it as work done is E.ds=Edscos@ when theta is 90 no work is done so considering a point charge to emit electric field in a radial field to its direction of motion @=90 and work done is 0 |dw:1372474806287:dw|
But Salini, that only holds good when you are doing the work IN AN EXTERNAL FIELD, here, the object (charge itself) has electric field, and we are talking about that field. So I agree with Carl that when you do positive work in CREATING electric field here, you also sort of do negative work in REMOVING the electric field at infinity.. and so the net work you do is zero!
well been away for a long time... i thought about it and figured that the work done is zero.but the reason i was looking for is that when we bring a charge somewhere electric field is created-so we may believe that we have done some work. But in reality,before bring the charge there,it was somewhere else.we just shifted the electric field associated around the charge from that region to a new region of space.