Understanding Delta PE: Why is it Equal to -W?

[MIA6]

1. Why delta PE=-W? In gravitation’s case, If we look at it numerically, PE= mg (b-a), W=F (b-a), then they are the same, where does the negative sign come from? Maybe that has to do with against gravitation or not? Vba=-Wba/q has the same concept. But here is one question: What minimum work must be done by an external force to bring a charge from a great distance away to a point from another charge. This is the general information, I didn’t put the numbers. Solution is W=q(Vb-Va), so why here there is no negative sign? because of the external force?
2. If the electric field is between two point charges, then are these two charges the source charges of the field? Or only one of them?

Thanks.

 

[atyy]

Well, I don't know about minus signs, I just choose them so the equations come out right. Anyway, it usually comes down to something in the definitions like "work done by the gravitational field" versus "work done against the gravitational field".

The electromagnetic force is an interaction, as is the gravitational force, so the gravitational field is generated by the masses of both particles. In the case where one particle is very much more massive than the other, we can neglect the motion of the larger mass, and consider the other as a "test mass". In that case, we say that the test mass moves in the gravitational field created by the larger mass.

 

[jtbell]

where does the negative sign come from?

The negative sign is necessary so that we can add potential energy and kinetic energy to get the total mechanical energy of the object. Imagine holding an object in your hand and moving it up or down in a gravitational field. The change in the object's KE equals the total work done on it, by gravity and by you:

$$\Delta KE = W_{grav} + W_{you}$$

$$\Delta KE = (- \Delta PE_{grav}) + W_{you}$$

$$\Delta KE + \Delta PE_{grav} = W_{you}$$

$$\Delta (KE + PE_{grav}) = W_{you}$$

The same argument holds for the work done by the electrical force, and electrical potential energy.

 

[vinay uppal]

Technically, potential energy of a conservative force is DEFINED to be the negative of the work done by the force between two points in space. It makes description of the work-energy theorem(for conservative forces) more convenient i.e in terms of the energy conservation law. You could always choose not to deal in potential energies, and use the work-energy theorem instead of the energy conservation law. The physics of a phenomenon will not change.

 

[MIA6]

The electromagnetic force is an interaction, as is the gravitational force, so the gravitational field is generated by the masses of both particles. In the case where one particle is very much more massive than the other, we can neglect the motion of the larger mass, and consider the other as a "test mass". In that case, we say that the test mass moves in the gravitational field created by the larger mass.

So what you mean in the case of the electric field is between two point charges, we consider one charge as test charge, the other one as source charge? depends on what though?

 

[atyy]

So what you mean in the case of the electric field is between two point charges, we consider one charge as test charge, the other one as source charge? depends on what though?

In the case of two point charges, if the smaller charge is much less than the bigger charge, so that the smaller charge does not affect the motion of the bigger charge, then we can consider the smaller charge to be a test charge.

 

[MIA6]

In the case of two point charges, if the smaller charge is much less than the bigger charge, so that the smaller charge does not affect the motion of the bigger charge, then we can consider the smaller charge to be a test charge.

What if two charges both carry for example 12 coulombs, then which is the source charge?

 

[atyy]

What if two charges both carry for example 12 coulombs, then which is the source charge?

If the problem states that one of the charges is fixed in space, then that is the source charge, and the other can be considered the test charge. Actually, the charge that cannot move can always be considered the source charge, even if its charge is smaller. The test charge is the one that is allowed to move.

If both charges are allowed to move at low velocities, then the concept of electric potential still applies. But in general, if all the charges involved move, then the problem cannot be solved using the concept of electric potential.

 

[MIA6]

ook. thanks.