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

$C=\frac{\varepsilon_{0} A}{d}$
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.
$C_{1}=\frac{K_{1} \varepsilon_{0}(A / 2)}{d}$
$=\frac{8}{2} \frac{\varepsilon_{0} A}{d}=\frac{4 \varepsilon_{0} A}{d}=4 C$
and $C_{2}=\frac{K_{2} \varepsilon_{0}(A / 2)}{d}$
$=\frac{4}{2} \frac{\varepsilon_{0} A}{d}=\frac{2 \varepsilon_{0} A}{d}=2 C$
Hence, their equivalent capacitance is
$C_{ eq }=C_{1}+C_{2}=4 C+2 C=6 C$
i.e. new capacitance will be six times the original