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Consider a parallel-plate capacitor in free space. A negatively charged point particle with initial velocity $v$ passes through the space between the pair of parallel plates (with an initial path perpendicular to the normal vector of the plates).

The point particle accelerates towards the positively charged plate but passes beyond the edge of the plate.

How is energy conserved, given that the capacitor does work on the particle by accelerating it in the direction towards the negatively charged plate?

EDIT: Was reminded by Art Brown that a negatively charged particle accelerates towards the positive plate.

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2 Answers 2

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This question is quite a common one for those first learning about capacitors.

First, let's remember that an electric field caused by stationary charges is conservative--this can easily be explained since a single charge creates a conservative field, and superposition of two conservative fields creates another conservative field.

So, the field generated by a floating capacitor has to be conservative. The universe isn't crazy, so it's probably us missing something? Are there any assumptions that we made while calculating the field? Yes, there are:

We assumed that the capacitor was infinite in size, and thus the field became uniform.

But, here, we are dealing with the edges of the capacitor. The field is not uniform here, it is more like (second half of image):

enter image description here

or:

enter image description here

When it comes back out, the x-component of the field will be against the velocity of the particle, slowing it down back to the initial speed. For example, for a positively charged particle, the trajectory is as follows:

enter image description here

The green indicates the force on the particle at various points. Once the particle exits, it is "pulled back". The net effect is that the speed stays the same but the direction does not. Perfectly in accordance with conservation of energy.

Ignoring fringe fields can lead to some interesting apparent paradoxi, like the origin of the force that pulls a dipole slab into a capacitor.

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@CrazyBuddy: I..don't quite get what you mean. Could you come to chat? –  Manishearth Dec 8 '12 at 7:52

The point charge $q$ moves in a potential field $\phi$ (generated by the capacitor), so the point charge has potential energy $U=q \phi$. It is accelerated by a force $\boldsymbol{F}$ along the gradient of that potential ($\boldsymbol{F}=q \boldsymbol{E}= -q \boldsymbol{\nabla} \phi$). For any such situation, you can show from the equation of motion ($\boldsymbol{F}=m\boldsymbol{a}$ in Newtonian dynamics) that the sum of the point charge's kinetic and potential energies is constant at all times.

In particular, if the potential field goes to 0 at infinity in every direction, a particle starting at infinity with speed $v$ ($v = $ magnitude of the velocity) that eventually escapes to infinity must end up with the same speed (assuming no collisions en route, as assumed by the problem).

(By the way, a negatively charged particle will be attracted to the positively charged plate.)

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thanks for the reminder in the last line of your answer! your answer's definitely mathematically beautiful but I felt that @Manishearth explained the reasons slightly better. I wish i could indicate both answers as the accepted answer though :) –  Vincent Tjeng Dec 9 '12 at 6:21
    
@VincentTjeng: No problem! You're welcome. –  Art Brown Dec 9 '12 at 6:27

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