Electrostatics Question 104
Question: The capacity of a parallel plate capacitor with no dielectric substance but with a separation of 0.4 cm is $ 2\mu F $ . The separation is reduced to half and it is filled with a dielectric substance of value 2.8. The final capacity of the capacitor is [CBSE PMT 2000]
Options:
A) $ 11.2\mu F $
B) $ 15.6\mu F $
C) $ 19.2\mu F $
D) $ 22.4\mu F $
Show Answer
Answer:
Correct Answer: A
Solution:
$ C=\frac{{\varepsilon _{0}}KA}{d} $
therefore $ \frac{C _{1}}{C _{2}}=\frac{K _{1}}{K _{2}}\times \frac{d _{2}}{d _{1}} $
$ \frac{2}{C _{2}}=\frac{1}{2.8}\times \frac{(0.4/2)}{(0.4)} $
therefore $ C _{2}=11.2\mu F $
