meldave00 said:
Russ,
Thanks for the reply. By the way, how do you quote other peoples words?
I haven't figured that out yet.
Ill put an extra character that you would not include in the example by using a '*'
To quote someone type...[*quote=whoeveryouarequoting]*Insert the persons text here*[*/quote]So the commands are [operation]thing to operate on [/operate] where the '[]' signify the operation and the '/' is the close for that operation.
To make text
bold type [*b]*Insert text here[*/b]
meldave00 said:
Any in response to your response. You said "Think about it like this...the plate3/plate2 “node” is "more negative" than plate4, but "less negative" than plate1."
What I'm saying is that there is a charge inbalance in the node between the two capacitors. So why is there not a voltage drop? I know that the E-Field inside a conductor is zero when in equilibrium.
Okay...Im going to give this a shot. Someone please correct me if I am wrong.
Here is what I believe happens...
Lets focus on the node in between the two capacitors. So we will look at both plates plus the wire conductor in between.
When the caps are charging we have one plate that basically deposits electrons onto the other plate via the wire conductor.
We still have the same
number of charges in the plate-wire-plate system, but the charges are now distributed differently than they were before.
On one plate we have a '+' charge because electrons were removed and on the other plate we have a '-' charge because electrons were "added".
But what about the wire in between? I would imagine that the free electrons in the wire are not evenly distributed along the wire but are displaced the exact opposite of the plates.
In other words, I would assume a higher electron density near the positive plate than near the negative plate. So an opposite electric field is created in the wire so the net field is zero.
Some might object at this point and ask why the electrons don't flow onto the positive plate and replace the charge deficiency there...I would answer that the electric field across the capacitor prevents this from occurring on both capacitors.
So we have four different fields in this situation.
Fields #1 and #2: The two fields across both capacitors repel and attract charges to and from the plates and create a potential energy between the plates. If we put a test charge in this region, it will flow from one point to another.
Fields #3 and #4: These two fields are set up by the charge distributions in between the two capacitors.
Field #3 is the field between the plates caused by the local distribution of extra electrons on one plate and missing electrons on the other plate.
Field #4 is the field created by a non-uniform distribution of charge in the wire conductor and cancels out field 3 such that a test charge placed anywhere in this region will experience no net electric field acting on it.So my short answer is that the reason there is no voltage drop is because the charge distribution in the wire cancels out the electric field created by the charge imbalance. Since there is then no net electric field, there is no voltage drop.
I hope that helps David...this is the best I can do for a quick answer. Someone please correct me if I am way off here.
Russ
****edited to add****
A thought just occurred to me about this...
Perhaps it is not the charge distribution in the
wire, but the charge distribution on the plates themselves that causes this opposite electric field.
Think about a single capacitor that is charged ...as one side becomes "negative" the other becomes "positive"...so the "positive" charged plate helps to hold the electrons in place on the "negative" charged plate.
Now disconnect the capacitor and think about the molecules and free electrons on both plates...
The negative plate has "extra" electrons on its surface that repel the remaining free electrons on the positive plate to the outside of the plate.
So if we looked at charge distribution from left to right going from negative plate to positive plate we get... '-' '+' '-'
Perhaps this distribution is what creates the opposing electric field bewteen the two capacitors.As a final thought maybe we can say that each individual situation is different depending on the value of the capacitors and how close they are together, but an opposing field is set up within the plate-wire-plate interface by the distribution of charges on the capacitor plates combined with the effect this has on the free electrons in the wire.
Does that make sense?
I would be surprised if the electron distribution in the wire in between the capacitors is uniform for all cases of two series capacitors.
Russ