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 $
Show Answer
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 $
BETA
