Ive got what seems to be a simple question, but I am completely confused. Im looking for the way to resolve this question, not neccessarily just the answer.
Basically the example im working with is: a 0.20 pF capacitor that has a voltage of 40 mV. How many more electroncs are on the negative plate compared to the positive plate?
2007-05-23 07:02:33 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Well a capacitor always has a charge Q on one plate and -Q on the other plate. With the data you're given you can easily find this Q:
C= Q/V so Q = CV.
But now you must think for a second.
When there is zero charge on the capacitor there is an equal number of electrons on both. But let's say the negative plate has a charge of -1e - that is it has 1 extra electron charge. Since the other plate has a charge of +1e, there are TWO more electrons on the negative plate as compared to the positive plate.
So when one plate has a charge -Q, it has 2Q more electrons than does the positve plate.
So in the end, solve for Q, and then multiply by 2 for your answer.
Make sense?
2007-05-23 07:15:34 · answer #1 · answered by Mikey C 2 · 0⤊ 0⤋
Well a .2 pf capacitor is so small it would be used in high freq . The 40 mv I have never Sean most would be a voltage of 25 v or more. U never want a cap in a ckt. that is closet to its breakdown voltage.
2007-05-23 15:43:35 · answer #2 · answered by JOHNNIE B 7 · 0⤊ 0⤋