A Parallel plate air capacitor, with the plate separation \(d\), has a capacitance of \(9\) pF. The space between the plates is now filled with two dielectrics: the first having \(\kappa_1 = 3\) and thickness \(d_1 = \frac{d}{3}\), while the second has \(\kappa_2 = 6\) and thickness \(d_2 = \frac{2d}{3}\). The capacitance of the capacitor is:
Solution:
The original air capacitor has a capacitance of \[ C_0 = \frac{\epsilon_0 A}{d} = 9 \text{ pF}. \] Filling the gap with two dielectrics in series, the effective capacitance is determined by: \[ \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2}, \] where \[ C_1 = \frac{\kappa_1 \epsilon_0 A}{d_1} \quad \text{and} \quad C_2 = \frac{\kappa_2 \epsilon_0 A}{d_2}. \] Substituting \( \kappa_1 = 3 \) and \( d_1 = \frac{d}{3} \): \[ C_1 = \frac{3 \epsilon_0 A}{d/3} = \frac{9\epsilon_0 A}{d} = 9 \times 9 \text{ pF} = 81 \text{ pF}. \] Similarly, for \( \kappa_2 = 6 \) and \( d_2 = \frac{2d}{3} \): \[ C_2 = \frac{6 \epsilon_0 A}{2d/3} = \frac{6 \epsilon_0 A \times 3}{2d} = \frac{18\epsilon_0 A}{2d} = \frac{9\epsilon_0 A}{d} = 81 \text{ pF}. \] Therefore, the overall capacitance becomes: \[ \frac{1}{C} = \frac{1}{81} + \frac{1}{81} = \frac{2}{81} \quad \Longrightarrow \quad C = \frac{81}{2} \text{ pF} = 40.5 \text{ pF}. \] Hence, option 3 is the correct answer.
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