Calculation of work done -- force confusion

[gracy]

Homework Statement

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

Homework Equations

$dW=$-$F^→.dr^→$

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.

 

[SteamKing]

Homework Statement

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
View attachment 89490

Homework Equations

$dW=$-$F^→.dr^→$

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.

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

 

[jbriggs444]

Homework Statement

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.

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.

Homework Equations

$dW=$-$F^→.dr^→$

The Attempt at a Solution

Now what I don't understand is why my book takes F is equal to $E^→.q$.

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.

I mean it moves by external force not the field force.Applied External force has nothing to do with force of electric field.

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.

 

[gracy]

the work done by an external force against that field in order to move the point charge?

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 $E^→$

 

[gracy]

the applied force (averaged over distance) must be equal and opposite to the field force

Why sir?

 

[jbriggs444]

Why sir?

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.

 

[jbriggs444]

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 $E^→$

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.

 

[gracy]

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.

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

 

[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?

 

[gracy]

Or only one?

only one will do!

 

[gracy]

then the external force must be essentially equal [and opposite] to the field force.

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

 

[jbriggs444]

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

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=f_field + f_external, if we know that f is zero and if f_field = 324.5 Newtons, what is f_external?

 

[gracy]

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

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

 

[jbriggs444]

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

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

 

[gracy]

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

The work done by the one and the work done by the other must add to zero.

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

 

[gracy]

If we assume that no kinetic energy is gained or lost then the total work must be zero.

The work done by the one and the work done by the other must add to zero.

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

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

 

[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?

 

[gracy]

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.

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

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

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?

 

[gracy]

what is fexternal?

-324.5 Newtons

 

[ehild]

Homework Statement

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 between p and s is dr.Calculate work done.

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.