Eng-Tips is the largest engineering community on the Internet

Intelligent Work Forums for Engineering Professionals

  • Congratulations waross on being selected by the Tek-Tips community for having the most helpful posts in the forums last week. Way to Go!

Capacitor Charging- Where are all the electrons? 11

Status
Not open for further replies.

WhMcC

Electrical
Aug 28, 2002
9
0
0
US
Greetings,

I am hoping that someone may have some insight here:

When a capacitor is charged, the electrons that are held in the electric field within the device must be physically distributed in some manner- how and where? Do the additional electrons fill normally vacant positions in the valence orbits of the material that comprises the plates of the capacitor? Or, is there some other mechanism by which the electrons are located and held in the field?

Thanks for your thoughts on this one!
 
Replies continue below

Recommended for you

Hi electricpete, that rule is just an approximation which only holds for the uniform parts of the plane, all you have tried to do is ignore the part that sinks your arguement.
 
cbarn,
We are talking about forces between plates in a capacitor and forces between charges on opposite side of a given plate. I have repeatedly identified the constant field assumption applies for plane geometry A>>d^2 which clearly applies in this context. Nothing hidden here. You have free choice to disbelieve the constant-field assumption if you are so-inclined but then of course you must also discard the formula C=epsilon*A/d which is based upon constant field assumption.
 
Hi electricpete-
I need to stop posting stuff so late. After reading them again my comments were pretty disjointed.
As you state, the energy required to incrementally add charge to a cap effectively increases with the square of the charge, due to the voltage increase in the cap. That part makes perfect sense so I will leave that alone.

But let's consider your car between two hills analogy. You correctly state that, intuitively, the car would eventually come to rest in the valley. However, if we used only the two energy equations that would seem to directly apply:

PE = mgh and KE = 0.5mv^2

and set the system in motion the car would never stop. Those two equations would swap energy perfectly indefinitely because there are no terms in the 'ideal' equations that produce any losses. What you would expect is that in a real test if you measure the speed of the car in the valley, the value of v would NOT be what you predict based on KE = PE = mgh. The difference would be the system losses.
In the capacitor question the voltages go to exactly what we expect them to. So where is the error from the 'ideal' equations that we would expect to see?
I think I may have just seen the error of my ways. We have the same discrepancy here as we see in mechanical systems with conservation of momentum. If you have two bodies of equal mass, one stationary and the other moving at velocity V and you model a collision between them with both bodies elastically deforming perfectly, conservation of momentum says you will end up with the two bodies moving at 0.5 * V. Momentum was conserved but the energy level of the system is cut in half. It's the same question; where's the energy lost to? So the thing I missed, as you state above, it the two base equations define fundamentally different quantities. You can't expect them to be equivalent.


 
Status
Not open for further replies.
Back
Top