A parallel plate capacitor with air between t | vedclass

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(C) The initial capacitance with air is $C = \frac{\varepsilon_0 A}{d} = 9 \, pF$.When the space is filled with two dielectrics of thickness $d_1 = d/

Vedclass · Accepted Answer

$40.5$

Vedclass · Answer

(C) The initial capacitance with air is $C = \frac{\varepsilon_0 A}{d} = 9 \, pF$.When the space is filled with two dielectrics of thickness $d_1 = d/3$ and $d_2 = 2d/3$,they act as two capacitors in series.The capacitance of the first part is $C_1 = \frac{K_1 \varepsilon_0 A}{d/3} = \frac{3 \varepsilon_0 A}{d/3} = 9 \frac{\varepsilon_0 A}{d} = 9 	imes 9 = 81 \, pF$.The capacitance of the second part is $C_2 = \frac{K_2 \varepsilon_0 A}{2d/3} = \frac{6 \varepsilon_0 A}{2d/3} = 9 \frac{\varepsilon_0 A}{d} = 9 	imes 9 = 81 \, pF$.Since they are in series,the equivalent capacitance $C'$ is given by $\frac{1}{C'} = \frac{1}{C_1} + \frac{1}{C_2}$.$\frac{1}{C'} = \frac{1}{81} + \frac{1}{81} = \frac{2}{81}$.$C' = \frac{81}{2} = 40.5 \, pF$.

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