Question

A parallel plate capacitor having cross-sectional area $A$ and separation $d$ has air in between the plates. Now an insulating slab of same area but thickness $d/2$ is inserted between the plates as shown in figure having dielectric constant $K(= 4)$. The ratio of new capacitance to its original capacitance will be,

• A. $2 : 1$
• B. $8 : 5$
• C. $6 : 5$
• D. $4 : 1$

Answer 1

Answer: (B)

$C_{a}=\frac{\in_{0} A}{d}$
$C_{k}=\frac{\in_{0} A}{d-t+\frac{t}{k}}$
$C_{k}=\frac{\in_{0} A}{d-\frac{d}{2}+\frac{d}{8}}$
$C_{k}=\frac{8}{5}\frac{\in_{0} A}{d}$
$C_{k}=\frac{8}{5}C_{a}$
$\frac{C_{k}}{C_{a}}=\frac{8}{5}$