# Capacitors

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#1
http://pmt.physicsandmathstutor.com/...Capacitors.pdf

Can someone explain 2ci.

I've looked at the mark scheme and the only bit I get is that the voltage in parallel for capacitors is the same.
Bit off topic but is the voltage in series for capacitors spread out or..?
0
5 years ago
#2
Pd across both capacitors is the same (5000v) because they are in parallel. Q = CV, Q of new capacitor = 1.2 x 10^-8 x 5000 = 6 x 10^-5 C. Charge left on ball is Q0 - (6 x 10^-5). Divide this by Q0 and you should get 1/1000 or 0.001. I always assume my working is wrong, so don't take my word for it.
0
5 years ago
#3
(Original post by Super199)
http://pmt.physicsandmathstutor.com/...Capacitors.pdf

Can someone explain 2ci.

I've looked at the mark scheme and the only bit I get is that the voltage in parallel for capacitors is the same.
Bit off topic but is the voltage in series for capacitors spread out or..?
the classic explanation for capacitors in series is given here
http://farside.ph.utexas.edu/teachin...de46.html#f6.2

with the the PD across the two capacitors in the question 6V the sum of the PD on the series capacitors will be 6V - in the trivial case of two identical capacitors it would be 3V across each series capacitor.
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