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

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(C) In the given arrangement,plate $Q$ is common to two capacitors connected in parallel,as both plates $P$ and $R$ are grounded (at $0 \,V$).Let $V$

Vedclass · Accepted Answer

$6.67$

Vedclass · Answer

(C) In the given arrangement,plate $Q$ is common to two capacitors connected in parallel,as both plates $P$ and $R$ are grounded (at $0 \,V$).Let $V$ be the potential of plate $Q$.The capacitance of the first capacitor (between $P$ and $Q$) is $C_1 = \frac{\varepsilon_0 A}{d}$.The capacitance of the second capacitor (between $Q$ and $R$) is $C_2 = \frac{\varepsilon_0 A}{2d}$.Since they are in parallel,the effective capacitance is $C_{	ext{eff}} = C_1 + C_2 = \frac{\varepsilon_0 A}{d} + \frac{\varepsilon_0 A}{2d} = \frac{3 \varepsilon_0 A}{2d}$.Given $q = 8.85 	imes 10^{-8} \,C$,$A = 2.0 \,m^2$,$d = 2 	imes 10^{-3} \,m$,and $\varepsilon_0 = 8.85 	imes 10^{-12} \,F/m$.Using $q = C_{	ext{eff}} V$,we have $V = \frac{q}{C_{	ext{eff}}} = \frac{q \cdot 2d}{3 \varepsilon_0 A}$.Substituting the values: $V = \frac{8.85 	imes 10^{-8} 	imes 2 	imes (2 	imes 10^{-3})}{3 	imes (8.85 	imes 10^{-12}) 	imes 2.0} = \frac{8.85 	imes 4 	imes 10^{-11}}{3 	imes 8.85 	imes 2 	imes 10^{-12}} = \frac{4 	imes 10}{3 	imes 2} = \frac{20}{3} \approx 6.67 \,V$.

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 (in $\,V$)

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