Electrostatics Question 103
Question: In a capacitor of capacitance $ 20\mu F $ , the distance between the plates is 2mm. If a dielectric slab of width 1mm and dielectric constant 2 is inserted between the plates, then the new capacitance is [BHU 2000]
Options:
A) $ 2\mu F $
B) $ 15.5\mu F $
C) $ 26.6\mu F $
D) $ 32\mu F $
Show Answer
Answer:
Correct Answer: C
Solution:
$ C=\frac{{\varepsilon _{0}}A}{d} $ and $ C’=\frac{{\varepsilon _{0}}A}{( d-t+\frac{t}{K} )} $
therefore $ \frac{C}{C’}=\frac{( d-t+\frac{t}{K} )}{d} $
$ \Rightarrow \frac{20}{C’}=\frac{( 2\times {{10}^{-3}}-1\times {{10}^{-3}}+\frac{1\times {{10}^{-3}}}{2} )}{2\times {{10}^{-3}}} $
therefore $ C’=26.6\mu F $
