Electrostatics Question 273
Question: Condenser $ A $ has a capacity of $ 15\mu F $ when it is filled with a medium of dielectric constant 15. Another condenser $ B $ has a capacity of $ 1\mu F $ with air between the plates. Both are charged separately by a battery of$ 100\ V $ . After charging, both are connected in parallel without the battery and the dielectric medium being removed. The common potential now is [MNR 1994]
Options:
A) $ \frac{3V}{K+2} $
B) $ \frac{3V}{K} $
C) $ \frac{V}{K+2} $
D) $ \frac{V}{K} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ q _{1}=CV, $
$ q _{2}=CV $
Now condenser of capacity C is filled with dielectric K, therefore C2 = KC As charge is conserved $ q _{1}+q _{2}=(C _{2}^{’}+C)V’ $
therefore $ V’=\frac{3CV}{(K+2)C}=\frac{3V}{K+2} $
BETA
