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

$C=\frac{{\epsilon }_{0}A}{d}$
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
$C_{eq}=C_1+C_2$
$=4C+2C=6C$
i.e., new capacitance will be six times of the original.