Electrostatics Question 804
Question: A parallel plate capacitor with air between the plates has a capacitance of 8 pF. Calculate the capacitance if the distance between the plates is reduced by half and the space between them is filled with a substance of dielectric constant. $ ( {\varepsilon _{r}}=6 ) $
Options:
A) 72 pF
B) 81 pF
C) 84 pF
D) 96 PF
Show Answer
Answer:
Correct Answer: D
Solution:
[d] Capacity of parallel plate capacitor
$ C=\frac{{\varepsilon _{r}}{\varepsilon _{0}}A}{d}\text{ }( \text{For air }{\varepsilon _{r}}=i ) $
So, $ \frac{{\varepsilon _{0}}A}{d}=8\times {{10}^{-12}} $
If $ d\to \frac{d}{2} $
and $ {\varepsilon _{r}}\to 6 $
then new capacitance $ C’=6\times \frac{{\varepsilon _{0}}A}{d/2}=12\frac{{\varepsilon _{0}}A}{d}=12\times 8pF=96pF $
BETA
