Question

Two parallel plates of area $A$ are separated by two different dielectric as shown in the figure. The net capacitance is

• A. $\frac{15 \epsilon _{0} A}{4 d}$
• B. $\frac{10 \epsilon _{0} A}{7 d}$
• C. $\frac{11 \epsilon _{0} A}{5 d}$
• D. $\frac{12 \epsilon _{0} A}{3 d}$

Answer 1

Answer: (A)

Since, we know,
Capacitance of a parallel plate capacitance,
$C=\frac{K \epsilon _{0} A}{d}$
Where, symbols have their usual meaning.
In the given figure,
$\frac{1}{C_{n e t}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}=\frac{1}{\frac{K_{1} \epsilon _{0} A}{d / 2}}+\frac{1}{\frac{K_{2} \epsilon _{0} A}{d / 2}}$
$=\frac{1}{\frac{3 \epsilon _{0} A}{d / 2}}+\frac{1}{\frac{5 \epsilon _{0} A}{d / 2}}$
$=\frac{d}{6 \epsilon _{0} A}+\frac{d}{10 \epsilon _{0} A}=\frac{d}{2 \epsilon _{0} A}\left[\frac{1}{3} + \frac{1}{5}\right]$
$=\frac{8 d}{15 \times 2 \times \epsilon _{0} A}=\frac{4 d}{15 \epsilon _{0} A}$
Thus, $C_{n e t}=\frac{15 \epsilon _{0} A}{4 d}$