# Calculation of work done -- force confusion

1. Sep 28, 2015

### gracy

1. The problem statement, all variables and given/known data
Consider a point charge Q placed at point O.Consider any point p in the field of the Q.Charge q is placed at point p.Charge q is then moved from point p to s the distance i=between p and s is dr.Calculate work done.
y

2. Relevant equations
$dW=$-$F^→.dr^→$

3. The attempt at a solution
Now what I don't understand is why my book takes F is equal to $E^→.q$
I mean it moves by external force not the field force.Applied External force has nothing to do with force of electric field.

2. Sep 28, 2015

### SteamKing

Staff Emeritus
Yes, but how do you know how much external force to apply to move the charge away from Q?

BTW, F = qE should be lurking somewhere in your chapter on electrostatics.

https://en.wikipedia.org/wiki/Electrostatics

3. Sep 28, 2015

### jbriggs444

Do you want to calculate the work done by the electric field on the point charge as it is moved? Or the work done by an external force against that field in order to move the point charge?

Assuming that the point charge starts and ends at rest then the Work Energy theorem has something to say. The work done by the one and the work done by the other must add to zero. The other will have the same magnitude and opposite sign.

So, up to sign, it does not matter which one you calculate. Pick the one that you can calculate from the information at hand.

The book has chosen to calculate the work done by the field. The force by the field on the charge is obviously given by the above formula.

It does have something to do with it. As above, the applied force (averaged over distance) must be equal and opposite to the field force.

4. Sep 28, 2015

### gracy

actually I don't know my book mentions this work done calculation while calculating voltage at point p.I don't know whether this work done is by electric field or any other force.And it does mention that F external is equal to - q $E^→$

Last edited: Sep 28, 2015
5. Sep 28, 2015

### gracy

Why sir?

6. Sep 28, 2015

### jbriggs444

By the work energy theorem. The integral of force times incremental distance adds up to work done. If we assume that no kinetic energy is gained or lost then the total work must be zero. So the integral of the field force over the path must be equal and opposite to the integral of the external force over the path. So their averages (over the path) must be equal and opposite.

7. Sep 28, 2015

### jbriggs444

If you want to move the charge slowly (i.e. without accelerating it significantly) or if the charge is of negligible mass (like an electron) then the external force must be essentially equal [and opposite] to the field force.

8. Sep 28, 2015

### gracy

or or and?I mean are these two essential conditions or only one of them would be fine for the external force to be equal [and opposite] to the field force

9. Sep 28, 2015

### jbriggs444

You tell me. In the equation f=ma, what two conditions could make f=0. Do both conditions need to apply? Or only one?

10. Sep 28, 2015

### gracy

only one will do!

11. Sep 28, 2015

### gracy

but then the force would become zero.how equal and opposite to the field force?

12. Sep 28, 2015

### jbriggs444

In the equation f=ma, f denotes the total force. It is the sum of all of the individual forces on the body.

If f=ffield + fexternal, if we know that f is zero and if ffield = 324.5 Newtons, what is fexternal?

13. Sep 28, 2015

### gracy

I think we should rather think of how the workdone becomes zero,Not force

14. Sep 28, 2015

### jbriggs444

The fact that you do not like the question you asked is no reason to object to the answer.

15. Sep 28, 2015

### gracy

You did not say/write that force should add up to zero.

16. Sep 28, 2015

### gracy

17. Sep 28, 2015

### jbriggs444

Please stop responding multiple times to the same posting. It is annoying. I am no longer certain what point you are trying to make. Work done is intimately related to force applied.

If the net force is always zero then the work done will clearly be zero regardless of the path traversed. That follows trivially from the fact that the integral of zero is zero.

If the work done is zero and the path traversed is of non-zero length then the net force will average to zero as long as we do an average over distance. That follows trivially from the definition of the suitable average.

So pick one -- force or work. Which one do you want to talk about? What question do you have about it?

18. Sep 29, 2015

### gracy

Why are you making individual force zero by taking "a" (acceleration)zero OR mass zero.Individual forces have to have certain magnitude(but equal and opposite)to make NET FORCE zero and as a result work done would be zero no matter how much the distance traveled is.So we are making the net force zero and that too by equal and opposite forces not by putting m=0
as far as acceleration =0 is concerned it is required here because the electron should not change it's kinetic energy thus not accelerate.It can increase it's potential energy.I am not getting why m≅0 is an important point here?

19. Sep 29, 2015

### gracy

-324.5 Newtons

20. Sep 29, 2015

### ehild

When speaking about "Work" we have to specify the force that does the work and the object on the work is done. So the question in the OP is meaningless, as it does not specify the force.
It can ask the work done by the electric field on the charge q while it moves from p to s. Or an external force can be specified and the work of that force can be asked.