Question

A parallel plate capacitor with air between the plates has a capacitance of $9pF$. The separation between its plates is $d$. The space between the plate is now partially filled with two dielectrics. One of the dielectric has dielectric constant $K_1=3$ and thickness $\frac{d}{3}$ while the other one has dielectric constant $K_2=6$ and thickness$\frac{d}{3}$. The remaining $\frac{d}{3}$ space is filled with air. The new capacitance is :

• A. $18pF$
• B. $45pF$
• C. $40.5pF$
• D. $20.25pF$

Answer 1

The correct option is A $18pF$

The effective capacitance of above combination is series combination of

$K_1\frac{{\epsilon }_{o}A3}{d},K_2\frac{{\epsilon }_{o}A3}{d}and\frac{{\epsilon }_{o}A3}{d}$

$\frac{1}{C_{ef}}=\frac{d}{3{\epsilon }_{0}A\times 3}+\frac{d}{6{\epsilon }_{0}A\times 3}+\frac{d}{3{\epsilon }_{0}A}$

$\frac{1}{C_{ef}}=\frac{d}{9{\epsilon }_{0}A}+\frac{d}{18{\epsilon }_{0}A}+\frac{d}{3{\epsilon }_{0}A}$

$C_{ef}=\frac{2\times {\epsilon }_{0}A}{d}$

$\Rightarrow C_{eff}=2\times 9pF$

$\Rightarrow C_{eff}=18pF$