# Does charge is also responsible for mass?

1. May 3, 2010

### astro2cosmos

according to newton's 2nd law of motion: a body is in remain its state unless an external force is applied to it. But i confuse if there is any charged body does it have any internal force?
suppose there are 2 bodies of same mass(m) but one have a charge of 10C & other is free of charge. then how much work should i done to move both bodies for a distance(d) ?
are both equal or not?

2. May 3, 2010

### tiny-tim

Hi astro2cosmos!
It depends what types of force are acting on the body.

If the forces include an electric or magnetic force, then their effect on the charged body will be different; if they don't, their effect will be the same.

3. May 3, 2010

### Starwanderer1

Hello astro2cosmos,

First of all, you have made a mistake in calling it the 2nd law, it is actually the 1st law..

You didnt mention if you consider these two events(moving two objects through a distance) independent ?

If you are an independence lover,then the answer would depend on the medium in which you are conducting the experiment..

For the 1st body(uncharged)life is simple. In the gravitational field of a mass M you would find work done to move 'm' through r to be GMm/r..

For the 2nd body-

in a medium that doesnt exchange charge(interact electromagnetically),work done would be the same if M is uncharged..

if M is charged then it would depend on the nature of charges on the bodies which decides if the work done against the electric field of M is +ve or-ve while gravitational work remains the same..

if the medium interacts in such a way that electric effects due to it & that due M are cancelled calculations would again give the same result as for the 1st mass.

Finally if these are moved together, the mutual gravitational interactions will play a part. Further if they are at a distance such that one can induce charges on the other the answer would vary again..

Thus the answer depends on a number of parameters concerning an external mass and the medium..

4. May 3, 2010

### HallsofIvy

Only charged bodies attract or repel each other. If one only one of bodies is charged, there will be no electro-magnetic force between them and so gravity will be the only relevant force.

Now when you say "move both bodies for a distance(c)", in which direction" Away from each other? Toward each other? Parallel to each other?

5. May 6, 2010

### astro2cosmos

all you are taking this question in anther way. my confusion is not by any em force or by gravitation force but it is that if i pull or push the bodies individually then for which body i would have to done more work against the resistance of body?

6. May 6, 2010

### xxChrisxx

That would be the first law.
EDIT: Nevermind someone already said that.

7. May 6, 2010

### ZapperZ

Staff Emeritus
You need to carefully and thoughtfully describe this, because this run-on sentence in itself is confusing.

If you push or pull on a body, you are doing work. If you pull on a body using a force F over a distance x, you're doing W amount of work. If you push on a body using the SAME force over the SAME distance, you're doing the identical amount of work. Is this what you are having a problem with?

Zz.

8. May 7, 2010

### astro2cosmos

you now understand little bit of my confusion. i mean that to move a fully uncharged body(m=2kg) for a distance of 1m, i'll do work of 2J. but if i move a fully charged body (assume any charge +ve or -ve) of 2kg for a distance of 1m, will i do the work of 2J?
does charging of body affect the magnitude of applied force for same distance?

9. May 7, 2010

### ZapperZ

Staff Emeritus
The nature of the force is irrelevant. It could be electromagnetic, gravitational, strong, etc. If you move a body with a mass using a force, you do work.

Zz.

10. May 7, 2010

### Staff: Mentor

W=f.d

Charge does not enter in to it.

11. May 7, 2010

### tiny-tim

If you move them across a completely smooth horizontal table, you'll do no work at all (the same in both cases).

If you move them across a rough horizontal table, you'll do work of friction times distance, the same in both cases.

If (in either case) there is also an electric field E, you will do extra work of E.d for the charged body.

12. May 7, 2010

### GRDixon

With regard to internal force, I suggest "The Feynman Lectures on Physics," V2, Sect. 28-4. The net self-force experienced by a charge is zero when its velocity is constant. However, when it accelerates and when da/dt<>0, the charge will experience an electric force in its own acceleration-induced electric field. This force must be counteracted by a driving agent if the acceleration is to remain as stipulated. With regard to two bodies, identical in every way except that one carries an excess electric charge, the charged body has electromagnetic mass, in addition to its mechanical mass, and it will take a greater applied force to accelerate it at a given rate. If the acceleration is constant (da/dt=0) then the impulse delivered by part of this force will pour out into the body's electromagnetic field. (It will pour back in from the fields if the charge is subsequently decelerated.) If both a and da/dt are nonzero, then the part of the agent force that counteracts da/dt will be manifest as radiant energy, always away from the charge and pouring into infinite space. This part of the total work done by the driving force cannot be recovered. Sometimes the part of the self-force ascribable to nonzero acceleration is called the inertial reaction force; whereas the part ascribable to nonzero da/dt is called the radiation reaction force. BTW, in Eq. 28.9 of the Feynman reference, the sign is wrong. The inertial reaction part of the total self force points opposite to the acceleration; the radiation reaction part points in the same direction as da/dt.