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Q.
A parallel plated capacitor has area $2 \,m^2$ separated by 3 dielectric slabs. Their relative permittivity is $2, 3, 6$ and thickness is $0.4 \,mm, 0.6\, mm, 1.2\, mm $, respectively. The capacitance is
COMEDKCOMEDK 2015Electrostatic Potential and Capacitance
Solution:
The given situation is equivalent to three capacitors $C _1, C _2$ and $C _3$ in series,
$
C_1=\frac{\varepsilon_0 K_1 A}{d_1}, C_2=\frac{\varepsilon_0 K_2 A}{d_2}, C_3=\frac{\varepsilon_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}{\varepsilon_0}\left(\frac{d_1}{K_1}+\frac{d_2}{K_2}+\frac{d_3}{K_3}\right)
$
Here, $A=2 m 2, K_1=2, K_2=3 ; K_3=6$
$d_1=0.4 mm =4 \times 10^{-4} m$
$d_2=0 \cdot 6 mm =6 \times 10^{-4} m$
$d_3=1 \cdot 2 mm =12 \times 10^{-4} m$
$\varepsilon_0=8 \cdot 85 \times 10^{-12} C ^2 N ^{-1} m ^{-2}$
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}{\varepsilon_0 A} \frac{d_1}{K_1} ; C=\frac{\varepsilon_0 A K_1}{3 d_1}
$
$
=\frac{8.85 \times 10^{-12} \times 2}{3 \times 2 \times 10^{-4}}=2.95 \times 10^{-8} F
$