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 $



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