Question

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

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

Answer 1

Answer: (B)

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 dielectrics, we have two capacitors of capacitance.
$C_1=\frac{K_1{\epsilon }_0(A/2)}{d}$
$=\frac{8}{2}\frac{{\epsilon }_{0}A}{d}=\frac{4{\epsilon }_{0}A}{d}=4C$
and $C_2=\frac{K_2{\epsilon }_0(A/2)}{d}$
$=\frac{4}{2}\frac{{\epsilon }_{0}A}{d}=\frac{2{\epsilon }_{0}A}{d}=2C$
Hence, their equivalent capacitance is
$C_{eq}=C_1+C_2=4C+2C=6C$
i.e. new capacitance will be six times the original