Question

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

• A. 12
• B. 6
• C. 8/3
• D. 3

Answer 1

Answer: (B)

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 dis 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 $C_2=\frac{{}^d\frac{K_2{\epsilon }_0(A/2)}{d}}{d_0}$
$\frac{4}{2}\frac{{\epsilon }_{0}A}{d}=\frac{2{\epsilon }_{0}A}{d}=2C$
Hence, their equivalent capacitance
$C=C_1+C_2$
$=4C+2C$
$=6C$
ie, new capacitance will be six time of the original.