Electrostatics Question 568
Question: Two capacitors of 3pF and 6pF are connected in series and a potential difference of 5000$ V $ is applied across the combination. They are then disconnected and reconnected in parallel. The potential between the plates is [MP PMT 1992]
Options:
A) 2250$ V $
B) 2222$ V $
C) $ 2.25\times 10^{6}V $
D) $ 1.1\times 10^{6}V $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \frac{1}{C}=\frac{1}{3}+\frac{1}{6} $
therefore $ C=2pF $
Total charge$ =2\times {{10}^{-12}}\times 5000={{10}^{-8}}C $
The new potential when the capacitors are connected in parallel is
$ V=\frac{2\times {{10}^{-8}}}{(3+6)\times {{10}^{-12}}}=2222V $
