A capacitor has a capacitance of 2.2 x 10^-8 F. In the charging process, electrons are removed from one plate and placed on the other plate. When the potential difference between the plates is 460 V, how many electrons have been transferred?
2007-09-05 11:40:13 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics
c=q/V
Find q
divide this number by the charge of electron 1.6*10^-19 C
you will get the number of electrons as an answer
2007-09-05 11:48:17 · answer #1 · answered by Snowflake 7 · 0⤊ 0⤋
Hmmm, the energy in a cap is CV^2/2 in joules, and one joule is 1 amp at 1 volt for 1 second...1 amp is 1 coulomb of electrons per second....bear with me, I'm trying to do this on the fly...oh wait, C=Q/V, so Q (charge) equals CV.
2.2 * 10^-8 *460V = 1012*10^-8 = 10.12 * 10^-6 coulombs.
BUT! That's the charge, plus and minus, so in getting there, only half that amount of electrons were transferred, right? So 5.06 microcoulombs. Final answer.
[REPOST] Ok, since ...
"A coulomb is then equal to exactly 6.241 509 629 152 65Ã10^18 elementary charges." [wikipedia, thanx], so
5.06 * 10^-6 * 6.241 509 629 152 65Ã10^18 =
31,582,038,723,512 electrons
2007-09-05 18:56:31 · answer #2 · answered by Gary H 6 · 0⤊ 0⤋