Question

The area of each plate of a parallel plate capacitor is $2m^2$. The space between the plates is filled with materials of dielectric constants 2, 3, and 6 and their thickness are 0.4 mm,0.6 mm and 1.2 mm respectively. The capacitance of the capacitor will be :

• A. $8.94\times {10}^{-4}F$
• B. $6.94\times {10}^{-7}F$
• C. $2.94\times {10}^{-8}F$
• D. ${10}^{-8}F$

Answer 1

The correct option is C $2.94\times {10}^{-8}F$

Here there are three capacitors in between the plates and they are in series.
$C_1=\frac{A{k}_1{ϵ}_0}{d_1}=\frac{2\times 2\times 8.854\times {10}^{-12}}{0.4\times {10}^{-3}}=8.854\times {10}^{-8}F$
$C_2=\frac{A{k}_2{ϵ}_0}{d_2}=\frac{2\times 3\times 8.854\times {10}^{-12}}{0.6\times {10}^{-3}}=8.854\times {10}^{-8}F$
$C_3=\frac{A{k}_3{ϵ}_0}{d_3}=\frac{2\times 6\times 8.854\times {10}^{-12}}{1.2\times {10}^{-3}}=8.854\times {10}^{-8}F$
thus, $C_1=C_2=C_3=C$
$\frac{1}{C_{eq}}=\frac{1}{C}+\frac{1}{C}+\frac{1}{C}=\frac{3}{C}$
$C_{eq}=\frac{C}{3}=\frac{8.854\times {10}^{-8}}{3}=2.94\times {10}^{-8}F$