Electrostatics Question 638
Question: In the circuit shown in ,$ C _{1}=6\mu F, $
$ C _{2}=3\mu F $ , and battery B = 20 V. The switch $ S _{1} $ is first closed. It is then opened, and $ S _{2} $ is closed. What is the final charge on$ C _{2} $ ?
Options:
A) $ 120\mu C $
B) $ 80\mu C $
C) $ 40\mu C $
D) $ 20\mu C $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] After closing $ S _{1} $
charge on $ C _{1} $ is$ q=6\times 20=120\mu C $ . Now, $ S _{1} $ is opened, on closing$ S _{2} $ ,
charge q will be distributed between $ C _{1} $ and $ C _{2} $ according to their copacitances.
So charge on $ C _{2} $ is $ q _{2}=\frac{C _{2}q}{C _{1}+C _{2}}=\frac{3\times 120}{3+6}=40\mu C $
BETA
