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 capactance 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
$C_{e q} =C_{1}+C_{2} $
$=4 C+2 C=6 C$
i.e., new capacitance will be six times of the original.