25 capacitors,each of capacitance 2 mu F,are | vedclass

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(D) In a series combination of $n$ identical capacitors each of capacitance $C$,the equivalent capacitance $C_{eq}$ is given by:$C_{eq} = \frac{C}{n}$

Vedclass · Accepted Answer

$8.0 	imes 10^{-5} C$

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(D) In a series combination of $n$ identical capacitors each of capacitance $C$,the equivalent capacitance $C_{eq}$ is given by:$C_{eq} = \frac{C}{n}$Given $n = 25$ and $C = 2 \mu F = 2 	imes 10^{-6} F$.$C_{eq} = \frac{2 	imes 10^{-6}}{25} F = 0.08 	imes 10^{-6} F = 8 	imes 10^{-8} F$.The potential difference applied is $V = 100 \ V$.The total charge $Q$ stored on the series combination is the same as the charge on each capacitor,given by:$Q = C_{eq} 	imes V$$Q = (8 	imes 10^{-8} F) 	imes (100 \ V)$$Q = 8 	imes 10^{-6} C$.

$25$ capacitors,each of capacitance $2 \mu F$,are connected in series to a battery of $100 \ V$. The total charge stored on the capacitors is

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