The energy isn't conserved here! Is there any thing wrong?

1. Jan 15, 2010

cometzir

The red dot represents a positively charged particle, the blue box represents a metal box, and the two dark green lines represent a capacitor, which has been charged. Each plate of the capacitor has a slit which will let the particle get through without touching the plates. And the capacitor doesn’t contact with the metal box either.

When the particle moves from A to B, it will be accelerated by the electric field. And its kinetic energy will increase.

When the particle is in the metal box, the electric field has no effects on the particle because of electrostatic shielding. So it consumes no energy to move the particle from B to A via C, D, and E.

Then, the particle will move from A to B through the electric field again, and its kinetic energy will increase again.

But in the whole process, it doesn’t need to produce work to keep the electric field, or to keep something else. So the energy except the kinetic energy of the particle doesn’t change. So the sum of energy increases.

Is there anything wrong?

(I am a foreigner, and my English is not very well,I hope you can understand what I write.)

Last edited by a moderator: Jan 15, 2010
2. Jan 15, 2010

espen180

I can't see your picture, but I'll try to explain.

When then electron is accelerated, the equipment excerts a torque on the earth, changing it's angular momentum. Energy is conserved, you were just not concidering a complete system.

3. Jan 15, 2010

pallidin

Yeah, that's called ramping. Done all the time. However, the energy required to induce ramping is greater than the energy of the output.

Last edited by a moderator: Jan 16, 2010
4. Jan 16, 2010

cometzir

So, it is different from the cyclotron, for the electric field passes through a diameter of the circle rather than a radius. So the equipment excerts two opposite torque on the earth per cycle, and doesn't chang its angular momentum. So the equipment can be regarded as a complete system, but the changing electric field will consume energy. Is it right?

But the electric field of the cyclotron is produced by the two metal box rather than another capacitor. Are there any differences between the electric field of the two equipment?

5. Jan 16, 2010

cometzir

Do you mean that the particle needs more energy to leave the electric field?

6. Jan 16, 2010

cometzir

Here is the picture

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7. Jan 16, 2010

Phrak

The picture was helpful. If you add up the electric potential around the loop, the net electrical potential is unfortunately zero, for those looking for free energy.

8. Jan 17, 2010

Staff: Mentor

There is a very sudden change in potential from outside the plates to between the plates. This sudden change in potential undoes the gain in kinetic energy that you get from accelerating between the plate. It is like rolling down a hill into slightly inclined wall of the same height.

9. Jan 17, 2010

uart

Hi cometzir. You understand that a metal box shields from electric fields becasue it is an equipotential surface (which is just another way of saying that all points in the metal are at the same voltage).

Now by definition the two plates on the charged capacitor must be at different potentials (from each other) therefore there must be an electric field between at least one of the capacitor plates and the box, and this is what creates the retarding electric field that DaleSpam refers to.

Hope that helps. :)

10. Jan 17, 2010

cometzir

So, there is another opposite electric field.But where is it? Is it in the metal box, or in the plate inside? If it is in the metal box, the electric field strength inside the box will not be zero. Is it possible?

11. Jan 18, 2010

uart

The opposing electric field is between the capacitor plates and the metal box. This is very straight forward.

12. Jan 18, 2010

cometzir

If the capacitor is out of the box, the opposing electric field can be like the red line in graph 1, but if the capacitor is on the border of the box like graph 2, or even in the box like graph 3, can you draw the opposing eletric field line?

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13. Jan 19, 2010

uart

Figure 1. Yes that's how it is, neither of the other options shown can work the way you want them to.

Figure 2. If the plates are in contact with the box then all you have is a short circuit and hence no E field. Reason : Box is a conductor, plates of capacitor are conductors, and they're touching. Doh, that should have been pretty obvious to you!

Remember that if you place any insulation layer at all between the box and the plates then you just end up with the first case (figure 1) above.

3. If the capacitor is inside the box (as per 3rd picture) then it will shield the E field from exiting the box and there will definitely still be a field inside the box.

14. Jan 19, 2010

Staff: Mentor

Also, one thing that you should keep in mind is that the approximation of the E-field in a capacitor as being entirely between the plates and uniform is just that, an approximation. In particular, it fails near the edges and on the outside the capacitor field looks similar to a dipole.

15. Jan 19, 2010

cometzir

I know it is impossible to let the equipment in the 1st and 3rd picture work. But in the 2nd pictrue, the pates can be smaller and aren't in contact with the box,as the following picture shows,if so ,where is the apposing E field?

For approximation, I know the E field isn't perfectly like the straight line.But between the two plates, I don't think there can be a apposing E field.
But I don't know whether the Electrostatic shielding is a approximation, in other words, whether there can be E field in the box if the plate isn't in the box.

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16. Jan 20, 2010

Staff: Mentor

If the plates are not in contact with the box then the opposing E-field is between the plates and the box. If the plates are in contact with the box then the capacitor discharges and there is no E-field. You cannot have both a gap in order to have an E-field and no gap in order to avoid the opposing E-field.

17. Jan 20, 2010

cometzir

It is true there can be an E field between the plate and the box, but in the second situation ,if the plate is not in contact with the box, the opposing E field will be perpendicular to the E field between the plate, even though the e field line isn't straight, it won't be "opposing".

Last edited: Jan 20, 2010
18. Jan 20, 2010

Staff: Mentor

I have no idea what you are saying here, but it does not really matter. In the Coulomb gauge for electrostatics we have:

$$\nabla^2V=-\frac{\rho}{\epsilon_0}$$

Which means that it does not matter how you arrange the charges $\rho$, you will always get a field that can be represented as the gradient of a scalar potential $V$. That automatically guarantees that the field is conservative. The fields simply don't behave the way you think.

19. Jan 20, 2010

uart

No the field wont just take the shortest (perpendicular) route to the box, it will be more like the situation in my attachment. (Which is extremely crude and not possible to show accurately in two dimensions but I hope it helps give you some sort of idea of a more realistic field pattern. Also I only "fixed up" the right hand portion.)

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20. Jan 20, 2010

cometzir

So the intensity of electric field in the box isn't zero even if the plate isn't in the box, and the electrostatic screen is just a approximation. Is it right?