A parallel plate capacitor with air between the plates has a capacitance of 12 pF. The separation between its plates is 'd' .The space between the plates is now filled with two dielectrics. One of the dielectrics has dielectric constant $k_{1} = 3$ and thickness $\frac{d}{3}$ while the other one has dielectric constant $k_{2} = 6$ and thickness $\frac{2 d}{3} .$ Capacitance of the

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54 pF

36 pF

18 pF

72 pF

answer is C.

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Detailed Solution

$The air filled capacitance , C = \frac{A ϵ_{0}}{d} = 12 pF$

According to question there are two capacitor with capacitances C₁ and C₂ and they are in series.

$Here, C_{1} = \frac{A k_{1} ϵ_{0}}{(d / 3)} = 9 \frac{A ϵ_{0}}{d} = 9 C as k_{1} = 3$

$and C_{2} = \frac{{Ak}_{2} ϵ_{0}}{(2 d / 3)} = 9 \frac{A ϵ_{0}}{d} = 9 C as k_{2} = 6$

$The equivalent capacitance, C_{eq} = \frac{C_{1} C_{2}}{C_{1} + C_{2}} = \frac{(9 C) (9 C)}{9 C + 9 C} = \frac{9}{2} C = (9 / 2) 12 = 54 pF$

$12 \times 3 + \frac{12 \times 3}{2} = 54 pF$

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