Super199
Badges: 20
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#1
Report Thread starter 5 years ago
#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
reply
Adam_1999
Badges: 8
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#2
Report 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
reply
Joinedup
Badges: 20
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#3
Report 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.
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest

How are you feeling about your results?

They're better than I expected (141)
40.52%
They're exactly what I expected (86)
24.71%
They're lower than I expected (121)
34.77%

Watched Threads

View All