Question

Is it just a coincidence that the word done to bring a charge from infinity to a particular point(which is the potential=kq/r) is the same as the work done by a constant force F=kq/r^2 in moving a charge by a distance r? Consider the charge to be a unit one.

Answer 1

Read carefully and try to understand yourself...

Answer 2

"constant force F=kq/r^2"

If force depends on r, it is not constant. I do not understand your point there.

Answer 3

It is just a coincidence that integrating 1/r² leads to -1/r.
When sign is changed to yield potential, it looks AS IF you had multiplied by r, but that's pure coincidence. That would not have happened if field had been in 1/r³

Answer 4

@Vincent-Lyon.Fr..here force is constant..i.e r is also constant.therefore workdone by this force will be also constant..i.e (k*q)/r. but for in electric potential expression r is variable..hence work_done by this constant force will be a particular value of electric potential not for all value of electric potential..electric potential is variable and it varies distance to distance from the electric field generator charge....

Answer 5

@ mani ji awesome question....

Answer 6

I am trying to get a physical interpretation out of this, but with no success. @Taufique, Are you saying that it is not a coincidence? I didn't understand you.
I've almost given up on this, taking it to be a co-incidence.

Answer 7

@Vincent-Lyon.Fr, the force can be constant if the source charge is very large(tending to infinity). The fact that we could have arrived at something simply by multiplying with r instead of integrating must mean something physically.
P.S. I decide not to give up! It shouldn't be a coincidence!

Answer 8

Of course, if force becomes constant, it's like PE in a uniform weight-field.

But then the 'constant' force does not vary like 1/r²

Answer 9

I don't want the force to vary, because I don't want to do integration. I've imagined something like this:
Suppose a source Q and a unit test charge placed r distance away. Now, suppose the charge Q gets larger and larger and ultimately becomes a very large sphere(but the charge remains same). While the Q is expanding, the test charge is moving away to maintain the constant distance r from Q. It is moving from the initial position to a very far point(infinity) just as the definition of potential demands. When the sphere becomes infinitely large, the test charge is infinitely far away from where it was initially. But the force between them was constant all along, though the test charge displaced.
Did you get the picture?
What would be the work done by Q in this case?

Answer 10

No idea! Sorry, but there are too many if's in your statement! I can't follow.

It's not I do not want to answer, but I think I've said all I could on the topic.

Answer 11

@Mani_Jha force can't remain constant between the charge Q and q in the process described by you in the last comment. Since force depends on the distance between the two charges and we measure the distance from the center of the two charges which is changing here. Since distance changes the work will be there. and you can calculate work by its simple formula

work done= (force).(displacement)

Answer 12

Thank you, I just realized that!

Answer 13

Do you also think that what i described in the question is just a coincidence?

Answer 14

No, it is not a coincidence.

Recall the work energy theorem; it says work done is equal to the change is energy. that's why the P.E. is equal to the work done.

Answer 15

In your past question ....24 faces participates in the flux .....