Two parallel plate capacitors of 8 mu F each | vedclass

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(B) Given,$C_1 = C_2 = 8 \mu F = 8 	imes 10^{-6} \ F$ and $V = 10 \ V$.Since $C_1$ and $C_2$ are connected in parallel,the initial equivalent capacit

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$120 \mu C$

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(B) Given,$C_1 = C_2 = 8 \mu F = 8 	imes 10^{-6} \ F$ and $V = 10 \ V$.Since $C_1$ and $C_2$ are connected in parallel,the initial equivalent capacitance is $C_{eq} = C_1 + C_2 = 16 \mu F = 1.6 	imes 10^{-5} \ F$.The initial total charge is $q_i = C_{eq} \cdot V = 1.6 	imes 10^{-5} 	imes 10 = 1.6 	imes 10^{-4} \ C = 160 \mu C$.We know that capacitance $C = \frac{\varepsilon_0 A}{d}$,so $C \propto \frac{1}{d}$.If the plate separation $d$ is reduced to $40 \%$ of its initial value,the new separation is $d' = 0.4d$. Thus,the new capacitance $C'$ becomes $C' = \frac{C}{0.4} = \frac{C}{2/5} = 2.5C = \frac{5}{2}C$.Wait,the problem states the separation is reduced $TO$ $40 \%$,meaning $d' = 0.4d$. So $C' = \frac{C}{0.4} = 2.5 	imes 8 \mu F = 20 \mu F$.New total capacitance $C_{eq}' = C' + C_2 = 20 \mu F + 8 \mu F = 28 \mu F$.New charge $q_f = C_{eq}' \cdot V = 28 \mu F 	imes 10 \ V = 280 \mu C$.Increase in charge $\Delta q = q_f - q_i = 280 \mu C - 160 \mu C = 120 \mu C$.

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

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