Question

A capacitor having capacitance $1\mu F$ with air is filled with two dielectrics as shown. How many times capacitance will increase ?

• A. $12$
• B. $6$
• C. $8/3$
• D. $3$

Answer 1

Answer: (B)

After filling with dielectrics the two capacitors will be in parallel order.
As shown, the two capacitors are connected in parallel. Initially the capacitance of capacitor $C=\frac{{\epsilon }_{0}A}{d}$
where, $A$ is area of each plate and $d$ is the separation between the plates.

After filling with two dielectrics, we have two capacitors of capacitance.
$C_1=K_1{\epsilon }_0\frac{\left(A/2\right)}{d}$
$=\frac{4}{2}\frac{{\epsilon }_{0}A}{d}$
$=\frac{4{\epsilon }_{0}A}{d}=4C$
and $C_2=\frac{K_2{\epsilon }_0\left(A/2\right)}{d}$
$=\frac{4}{2}\frac{{\epsilon }_{0}A}{d}$
$=\frac{2{\epsilon }_{0}A}{d}=2C$
Hence,their equivalent capacitance
$C^{\prime }=C_1+C_2=4C+2C$
$=6C$
i.e., new capacitance will be six times of the original.