A parallel plate capacitor with air between the plates has a capacitance of 9pF. 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 k₁= 3 and thickness d/3 while the other one has dielectric constant k₂ = 6 and thickness 2d/3. Capacitance of the capacitor is now

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40.5pF

1.8pF

45pF

20.25pF

answer is C.

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Detailed Solution

$\begin{array}{l} C_{0} = \frac{{Aε}_{0}}{d} = 9 pF; C_{1} = 3 k_{1} C_{0} = 3 \times 3 \times 9 = 81 pF \\ C_{2} = \frac{3}{2} k_{2} C_{0} = \frac{3}{2} 6 \times 9 = 81 pF \end{array}$
C₁ and C₂ are in series
$C_{1} = C_{2} = \frac{81}{2} = 40 .5 pF$