Confused about work done on charge

[Pau Hernandez]

TL;DR

How does work done affect velocity of charges?

Hello people,

super confused here. In electrostatics we define the work done on a point charge Q to move it from A to B as $$W = Q \cdot \int_A^B \vec{E} \cdot d\vec{s}$$ .
Where exactly does the energy go here? For instance when a positive charge Q is moved in the direction of the electric field lines, the field does work on the charge. I assume the field then loses this energy? Is the charge at rest or does it gain kinetic energy? And this is where it gets confusing. How can the charge be at rest when the coulomb force $$\vec{F}=Q \cdot \vec{E}$$ keeps accelerating it? Because in class we also derived that when an electron is exposed to a constant electric field and the electron moves through a potential difference of V, the change in kinetic energy of the electron will be: $$\Delta E_{kin} = q \cdot V$$ , where q is the charge of the electron.

 

[Ibix]

TL;DR Summary: How does work done affect velocity of charges?

Where exactly does the energy go here?

There's a close analogy with lifting an object in a gravitational field here. You have to apply a force to keep the object in a fixed position, you have to apply more force to start it moving (which translates to an increase in kinetic energy and the potential energy starting to increase), you can keep applying a force during the movement, and you reduce the force to let the object stop, but you still need to apply a force to keep it stopped. The forces keeping it stopped do no work, though.

So you are correct, the energy goes into (or comes from) the field. You will also add kinetic energy to the object and that will cost you work, but you can get it back when the object slows back down to a stop.

Mathematically, you apply a force $\vec F$ and the field applies a force $q\vec E$. The work you do on the particle is $\int \vec F\cdot d\vec s$ and the work the field does is $q\int\vec E\cdot d\vec s$. In general, the particle will gain kinetic energy $\Delta E=\int\vec F\cdot d\vec s+q\int\vec E\cdot d\vec s$. In the special case that the particle starts and finishes the motion at the same speed (e.g. at rest), $\Delta E=0$ and the two integrals must be equal but with opposite sign: the work you did on the object was the negative of the work the field did, so the net effect was to transfer energy between you and the field. You'll still have to apply a force before and after the motion, but since $ds=0$, the energy transfer is zero.

 

[Pau Hernandez]

Oh yes. Thats a good explanation. Funny how the simple things are not so simple after all. Thanks!

 

[Pau Hernandez]

One more question Mr. Ibix. Do you know of any good introductory book? Because in my book, it was never mentioned that there is some mechanism that actually has to keep the charge in place. Thats where the confusion about the energy transfer came from.

 

[PeroK]

One more question Mr. Ibix. Do you know of any good introductory book? Because in my book, it was never mentioned that there is some mechanism that actually has to keep the charge in place. Thats where the confusion about the energy transfer came from.

Griffiths' book on EM is not elementary, but he's very good, IMO, at pointing out things like this. On this point, for example, he says:

"As you bring in each charge, nail it down in its final position, so it doesn't move when you bring in the next charge."

 

[weirdoguy]

but he's very good, IMO, at

At everything I would say, great book

 

[tech99]

Maybe the moving charge stores energy in a magnetic field.

 

[cianfa72]

There's a close analogy with lifting an object in a gravitational field here. You have to apply a force to keep the object in a fixed position, you have to apply more force to start it moving (which translates to an increase in kinetic energy and the potential energy starting to increase)

Just a point is worth to be highlighted: when you lift an object without changing its kinetic energy, the potential energy starts increasing. But...what is the potential energy really ? It is associated to the overall configuration of bodies within the field. So, strictly speaking, the potential energy is not localized in the lifted object: it is a property of the "object within the field" system.

 

[cianfa72]

So, strictly speaking, the potential energy is not localized in the lifted object: it is a property of the "object within the field" system.

I would add that, in the realm of classic electrodynamics, one typically starts from a given spatial configuration of charges sources of the electric field E. Then when you "lift" a new charge into it, the underlying assumption is that the charges sources of the field are nailed down in their positions (this is basically the same as assuming a very small new charge not affecting the field's sources). Now, when you are done lifting the new charge, you will get a new system of charges/sources with a (new) associated electric field distribution. The main point is that the potential energy of such a new charges's configuration can be evaluated by integrating the square of the (new) electric field distribution over the whole space (actually it is enough to integrate it over the region the electric field doesn't vanish) ! In other words one can think of the potential energy of the system of charges as associated with the field itself.