Three parallel plates (each of area A) are separated by distances $d_1$ and $d_2$ respectively. The in-between spaces are filled with dielectrics of relative permittivity $ε_1$ and $ε_2$. The permittivity of free space is $ε_0$. The capacitance of the combination is

$ε_0 ε_1 ε_2 A/ (ε_2 d_1 + ε_1 d_2)$

The correct answer is Option (2) → $ε_0 ε_1 ε_2 A/ (ε_2 d_1 + ε_1 d_2)$

For capacitors in series, the equivalent capacitance is given by:

$\frac{1}{C_\text{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$

Capacitance of a parallel-plate capacitor with dielectric:

$C = \frac{\epsilon A}{d} = \frac{\epsilon_0 \epsilon_r A}{d}$

Here:

$C_1 = \frac{\epsilon_0 \epsilon_1 A}{d_1}$

$C_2 = \frac{\epsilon_0 \epsilon_2 A}{d_2}$

Series combination:

$\frac{1}{C_\text{eq}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{d_1}{\epsilon_0 \epsilon_1 A} + \frac{d_2}{\epsilon_0 \epsilon_2 A} = \frac{d_1}{\epsilon_0 \epsilon_1 A} + \frac{d_2}{\epsilon_0 \epsilon_2 A}$

So, the equivalent capacitance:

$C_\text{eq} = \frac{\epsilon_0 A}{\frac{d_1}{\epsilon_1} + \frac{d_2}{\epsilon_2}}$

Final Answer: $C_\text{eq} = \frac{\epsilon_0 A}{\frac{d_1}{\epsilon_1} + \frac{d_2}{\epsilon_2}}$