A 500,mu F capacitor is charged at a steady r | vedclass

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(D) The charge $Q$ on a capacitor is given by the formula $Q = C V$.Since the capacitor is charged at a steady rate,the rate of change of charge is $\

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$50$

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(D) The charge $Q$ on a capacitor is given by the formula $Q = C V$.Since the capacitor is charged at a steady rate,the rate of change of charge is $\frac{dQ}{dt} = I = 100\,\mu C/s = 100 	imes 10^{-6}\,C/s$.The capacitance $C = 500\,\mu F = 500 	imes 10^{-6}\,F$.We want to find the time $\Delta t$ required to reach a potential difference $V = 10\,V$.Using the relation $Q = C V$,the total charge accumulated is $Q = (500 	imes 10^{-6}\,F) 	imes (10\,V) = 5000 	imes 10^{-6}\,C$.Since the charge is supplied at a constant rate,$Q = I 	imes \Delta t$.Therefore,$\Delta t = \frac{Q}{I} = \frac{5000 	imes 10^{-6}\,C}{100 	imes 10^{-6}\,C/s} = 50\,s$.

$A$ $500\,\mu F$ capacitor is charged at a steady rate of $100\,\mu C/s$. The potential difference across the capacitor will be $10\,V$ after an interval of.....$s$.

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