A thin metallic partition of negligible thick | vedclass

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Question

(A) $1$. The total volume of liquid filled in each compartment after $t = 5 \ s$ is: Volume $V_1 = (	ext{flow rate}) 	imes t = 2 \ m^3/s 	imes 5 \

Vedclass · Accepted Answer

$8.85 	imes 10^{-11} \ F$

Vedclass · Answer

(A) $1$. The total volume of liquid filled in each compartment after $t = 5 \ s$ is: Volume $V_1 = (	ext{flow rate}) 	imes t = 2 \ m^3/s 	imes 5 \ s = 10 \ m^3$. Volume $V_2 = 1 \ m^3/s 	imes 5 \ s = 5 \ m^3$. $2$. The height of the container is $10 \ m$. The area of the base of each compartment is $1 \ m 	imes 1 \ m = 1 \ m^2$. Height of liquid in the first compartment $h_1 = V_1 / 	ext{Area} = 10 / 1 = 10 \ m$. Height of liquid in the second compartment $h_2 = V_2 / 	ext{Area} = 5 / 1 = 5 \ m$. $3$. The system forms a capacitor $C_1$ in series with a parallel combination of $C_2$ and $C_3$. $C_1 = \frac{K_1 \varepsilon_0 A_1}{d_1} = \frac{3 \varepsilon_0 (10 	imes 1)}{1} = 30 \varepsilon_0$. $C_2 = \frac{K_2 \varepsilon_0 A_2}{d_2} = \frac{1 \varepsilon_0 (5 	imes 1)}{1} = 5 \varepsilon_0$. $C_3 = \frac{K_3 \varepsilon_0 A_3}{d_3} = \frac{2 \varepsilon_0 (5 	imes 1)}{1} = 10 \varepsilon_0$. $4$. The equivalent capacitance $C_{eq}$ is: $C_{eq} = \frac{C_1 (C_2 + C_3)}{C_1 + C_2 + C_3} = \frac{30 \varepsilon_0 (5 \varepsilon_0 + 10 \varepsilon_0)}{30 \varepsilon_0 + 5 \varepsilon_0 + 10 \varepsilon_0} = \frac{30 \varepsilon_0 	imes 15 \varepsilon_0}{45 \varepsilon_0} = 10 \varepsilon_0$. $C_{eq} = 10 	imes 8.85 	imes 10^{-12} \ F = 8.85 	imes 10^{-11} \ F$.

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