E

#### Email Invalid

- Jan 1, 1970

- 0

series.

Example:

C1 C2

----| |----| |----

a b c

Let's say that at time t=0, ideal voltage sources are connected to

nodes a,b and c such that Va = 2V, Vb = 0V, Vc = 0V.

Then the "charge on C1" would be Q1=C1 * 2.

However, I thought that when we say that a capacitor stores charge

Q=CV, what it really means is that the more positive plate has +Q and

the more negative plate has -Q charge, right?

It is my understanding that the "left" side of C1 would have charge

+Q1 while the "right" side would have charge -Q1. But since VC2 = 0V,

then C2 stores no charge and so the "left" side of C1 should have 0

charge and the "right" side should have 0 charge.

First of all, this seems confusing because how could node b have -Q1

charge and 0 charge at the same time??? If the "right" plate of C1

has -Q1 charge, how could the "left" plate of C2 have 0 charge

simultaneously? Wouldn't node b have a total net charge of -Q1 so that

the "left" plate of C2 would also have charge -Q1. But that doesn't

make sense either since there is no voltage across C2 so C2 should

have no charge on it whatsoever. I am confused!

Okay, now let's say at t=t1, we remove the 0V voltage source on node b

so that it now "floats" but keep Va=2V and Vc=0V Then what is the

voltage and charge on node b?

What if we removed all the voltage sources, what then?

I guess I don't really understand Q=CV with a solid understanding.