Four metallic plates A, B, C and D of the sam | vedclass

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(C) Let the capacitance of each pair of plates without a dielectric be $C = \frac{\epsilon_0 A}{d}$.There are three capacitors formed by the plates: $

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$\frac{4}{5} C$

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(C) Let the capacitance of each pair of plates without a dielectric be $C = \frac{\epsilon_0 A}{d}$.There are three capacitors formed by the plates: $C_1$ between $A$ and $B$,$C_2$ between $B$ and $C$,and $C_3$ between $C$ and $D$.$C_1 = C$ (air between $A$ and $B$).$C_2 = K C = 2C$ (dielectric $K=2$ between $B$ and $C$).$C_3 = K C = 2C$ (dielectric $K=2$ between $C$ and $D$).Plates $B$ and $D$ are connected together. Thus,$C_2$ and $C_3$ are in parallel between points $B$ and $C$. The equivalent capacitance of this parallel combination is $C_{23} = C_2 + C_3 = 2C + 2C = 4C$.Now,$C_1$ is in series with the combination $C_{23}$ between points $A$ and $C$. However,looking at the circuit,$A$ is connected to one plate of $C_1$,and the other plate $B$ is connected to $C_2$ and $C_3$. Since $B$ and $D$ are connected,the effective capacitance between $A$ and $C$ is the series combination of $C_1$ and $C_{23}$.$C_{eq} = \frac{C_1 	imes C_{23}}{C_1 + C_{23}} = \frac{C 	imes 4C}{C + 4C} = \frac{4C^2}{5C} = \frac{4}{5} C$.

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