Question

A parallel plated capacitor has area $2m^2$ separated by 3 dielectric slabs. Their relative permittivity is $2,3,6$ and thickness is $0.4mm,0.6mm,1.2mm$, respectively. The capacitance is

• A. $5\times 10^{-8}Farad$
• B. $11\times 10^{-8}Farad$
• C. $2.95\times 10^{-8}Farad$
• D. $10\times 10^{-8}Farad$

Answer 1

Answer: (C)

The given situation is equivalent to three capacitors $C_1,C_2$ and $C_3$ in series,
$C_1=\frac{{\epsilon }_0{K}_{1}A}{d_1},C_2=\frac{{\epsilon }_0{K}_{2}A}{d_2},C_3=\frac{{\epsilon }_0{K}_{3}A}{d_3}$
Equivalent.capacitance (C) is given by
$\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}=\frac{1}{{\epsilon }_0}\left(\frac{d_1}{K_1}+\frac{d_2}{K_2}+\frac{d_3}{K_3}\right)$
Here, $A=2m2,K_1=2,K_2=3;K_3=6$
$d_1=0.4mm=4\times 10^{-4}m$
$d_2=0\cdot 6mm=6\times 10^{-4}m$
$d_3=1\cdot 2mm=12\times 10^{-4}m$
${\epsilon }_0=8\cdot 85\times 10^{-12}{C}^2{N}^{-1}{m}^{-2}$
Also, $\frac{d_1}{K_1}=\frac{d_2}{K_2}=\frac{d_3}{K_3}=2\times 10^{-4}m$
$\therefore \frac{1}{C}=\frac{3}{{\epsilon }_{0}A}\frac{d_1}{K_1};C=\frac{{\epsilon }_{0}A{K}_1}{3d_1}$
$=\frac{8.85\times 10^{-12}\times 2}{3\times 2\times 10^{-4}}=2.95\times 10^{-8}F$