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 \( \varepsilon_1 \) and \( \varepsilon_2 \). The permittivity of free space is \( \varepsilon_0 \). The capacitance of the combination is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \varepsilon_0 \varepsilon_1 \varepsilon_2 A / (\varepsilon_2 d_1 + \varepsilon_1 d_2) \)
The three plates form two capacitors, one in each gap, and the outer plates are the terminals of the combination. The same charge sits on both, so they are in series. A capacitor filled with a dielectric of relative permittivity \( \varepsilon_r \) has \( C = \varepsilon_0\varepsilon_r A/d \).
\( \displaystyle C_1 = \frac{\varepsilon_0 \varepsilon_1 A}{d_1}, \qquad C_2 = \frac{\varepsilon_0 \varepsilon_2 A}{d_2} \)
\( \displaystyle \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{d_1}{\varepsilon_0 \varepsilon_1 A} + \frac{d_2}{\varepsilon_0 \varepsilon_2 A} = \frac{\varepsilon_2 d_1 + \varepsilon_1 d_2}{\varepsilon_0 \varepsilon_1 \varepsilon_2 A} \)
\( \displaystyle C = \frac{\varepsilon_0 \varepsilon_1 \varepsilon_2 A}{\varepsilon_2 d_1 + \varepsilon_1 d_2} \)