Electrostatics Question 562

Question: A condenser having a capacity of 6mF is charged to 100 V and is then joined to an uncharged condenser of $ 14\mu F $ and then removed. The ratio of the charges on 6mF and 14mF and the potential of 6mF will be [MP PMT 1991]

Options:

A) $ \frac{6}{14} $ and $ 50\ volt $

B) $ \frac{14}{6} $ and $ 30\ volt $

C) $ \frac{6}{14} $ and $ 30\ volt $

D) $ \frac{14}{6} $ and 0$ volt $

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Answer:

Correct Answer: C

Solution:

Let $ q _{1},q _{2} $ be the charges on two condensers

$ V=\frac{q _{1}}{6}=\frac{q _{2}}{14} $

therefore $ \frac{q _{1}}{q _{2}}=\frac{6}{14}=\frac{3}{7} $

Also $ q _{1}+q _{2}=600 $

therefore $ q _{1}+\frac{14}{6}q _{1}=600 $

therefore $ q _{1}=\frac{600}{20}\times 6 $ $ V=\frac{q _{1}}{6}=\frac{600}{20}=30volt $



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