In the diagram,the area of each plate is A = | vedclass

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(C) The system consists of two capacitors in parallel connected to plate '$Q$'.One capacitor is formed by plates '$P$' and '$Q$' with separation $d$,a

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$\frac{20}{3} \,V$

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(C) The system consists of two capacitors in parallel connected to plate '$Q$'.One capacitor is formed by plates '$P$' and '$Q$' with separation $d$,and the other by plates '$Q$' and '$R$' with separation $2d$.The capacitance of a parallel plate capacitor is given by $C = \frac{\epsilon_0 A}{d}$.Let $C_1$ be the capacitance between '$P$' and '$Q$': $C_1 = \frac{\epsilon_0 A}{d}$.Let $C_2$ be the capacitance between '$Q$' and '$R$': $C_2 = \frac{\epsilon_0 A}{2d} = \frac{C_1}{2}$.Since both plates '$P$' and '$R$' are grounded,the potential of both is $0 \,V$. Thus,the capacitors are in parallel with respect to plate '$Q$'.The equivalent capacitance is $C_{eq} = C_1 + C_2 = C_1 + \frac{C_1}{2} = \frac{3}{2} C_1$.Substituting the values: $C_1 = \frac{8.85 	imes 10^{-12} 	imes 2}{2 	imes 10^{-3}} = 8.85 	imes 10^{-9} \,F$.$C_{eq} = \frac{3}{2} 	imes 8.85 	imes 10^{-9} = 1.3275 	imes 10^{-8} \,F$.The potential of plate '$Q$' is $V = \frac{q}{C_{eq}} = \frac{8.85 	imes 10^{-8}}{1.3275 	imes 10^{-8}} = \frac{8.85}{1.3275} = \frac{8.85}{1.5 	imes 8.85} 	imes 10 = \frac{20}{3} \,V$.

In the diagram,the area of each plate is $A = 2 \,m^2$ and $d = 2 \times 10^{-3} \,m$. An electric charge of $q = 8.85 \times 10^{-8} \,C$ is given to the plate '$Q$'. Plates '$P$' and '$R$' are connected to the ground. Then the potential of plate '$Q$' is:

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