Two parallel plate capacitors $8 \mu \mathrm{F}$ each are connected in parallel to a $10 \mathrm{~V}$ battery. The plate separation in one of the capacitor is reduced to $40 \%$ of its initial value. The increase in the total charge stored on the capacitors is

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Aicharya · Accepted Answer

Given, $ \begin{aligned} C_1 & =C_2=8 \mu \mathrm{F}=8 	imes 10^{-6} \mathrm{~F} \ V & =10 \mathrm{~V} \end{aligned} $ Since, $C_1$ and $C_2$ are connected in parallel. Equivalent capacitance, $ \begin{aligned} C_{	ext {eq }}=C_1+C_2= & 8 	imes 10^{-6}+8 	imes 10^{-6}=1.6 	imes 10^{-5} \mathrm{~F} \ 	ext { Total charge, } q_i & =C_{\mathrm{eq}} \cdot V \ & =1.6 	imes 10^{-5} 	imes 10 \ & =1.6 	imes 10^{-4} \mathrm{C} \end{aligned} $ We know that, capacitance $C=\frac{\varepsilon_0 A}{d}$ $ \begin{array}{ll} \Rightarrow & C \propto \frac{1}{d} \ \Rightarrow & \frac{C^{\prime}}{C^{\prime \prime}}=\frac{d^{\prime \prime}}{d^{\prime}}=\frac{\left(d^{\prime}-40 \% 	ext { of } d^{\prime}ight)}{d^{\prime}}=\frac{3 / 5 d^{\prime}}{d^{\prime}} \end{array} $ $ \begin{aligned} & \Rightarrow \quad \frac{C^{\prime}}{C^{\prime \prime}}=\frac{3}{5} \ & \Rightarrow \quad C^{\prime \prime}=\frac{5}{3} C^{\prime} \ & =\frac{5}{3} 	imes 8 	imes 10^{-6}=\frac{40}{3} 	imes 10^{-6} \mathrm{C} \ & \end{aligned} $ New capacitance, $C^{\prime \prime \prime}=C^{\prime \prime}+C_2$ $ \begin{aligned} & =\frac{40}{3} 	imes 10^{-6}+8 	imes 10^{-6} \ & =\frac{64}{3} 	imes 10^{-6} \mathrm{~F} \end{aligned} $ Now value of total charge, $q_f=C^{\prime \prime \prime} 	imes V$ $ \begin{aligned} & =\frac{64}{3} 	imes 10^{-6} 	imes 10 \ & =\frac{64}{3} 	imes 10^{-5} \mathrm{C} \end{aligned} $ Increased value of charge, $ \begin{aligned} \Delta q & =q_f-q_i \ & =\frac{64}{3} 	imes 10^{-5}-1.6 	imes 10^{-4} \ & =\frac{160}{3} \mu \mathrm{C} \end{aligned} $

Two parallel plate capacitors $8 \mu \mathrm{F}$ each are connected in parallel to a $10 \mathrm{~V}$…

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