See the diagram. The area of each plate is 2. | vedclass

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(C) The system consists of two capacitors connected in parallel between plate $Q$ and the grounded plates $P$ and $R$.The capacitance of the first cap

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$6.67$

Vedclass · Answer

(C) The system consists of two capacitors connected in parallel between plate $Q$ and the grounded plates $P$ and $R$.The capacitance of the first capacitor (between $Q$ and $P$) is $C_1 = \frac{\epsilon_0 A}{d}$.The capacitance of the second capacitor (between $Q$ and $R$) is $C_2 = \frac{\epsilon_0 A}{2d}$.Since both capacitors are connected in parallel,the equivalent capacitance is $C_{eff} = C_1 + C_2 = \frac{\epsilon_0 A}{d} + \frac{\epsilon_0 A}{2d} = \frac{3 \epsilon_0 A}{2d}$.Given $A = 2.0\ m^2$,$d = 2 	imes 10^{-3}\ m$,and $\epsilon_0 = 8.85 	imes 10^{-12}\ F/m$,we have:$C_{eff} = \frac{3 	imes 8.85 	imes 10^{-12} 	imes 2.0}{2 	imes 2 	imes 10^{-3}} = \frac{3 	imes 8.85 	imes 10^{-12}}{2 	imes 10^{-3}} = 1.5 	imes 8.85 	imes 10^{-9}\ F$.The potential $V$ of plate $Q$ is given by $V = \frac{q}{C_{eff}}$.Substituting the values: $V = \frac{8.85 	imes 10^{-8}}{1.5 	imes 8.85 	imes 10^{-9}} = \frac{10^{-8}}{1.5 	imes 10^{-9}} = \frac{10}{1.5} = 6.67\ V$.

See the diagram. The area of each plate is $2.0\ m^2$ and $d = 2 \times 10^{-3}\ m$. $A$ charge of $8.85 \times 10^{-8}\ C$ is given to plate $Q$. Then the potential of $Q$ becomes......$V$.

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