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If a capacitor of capacitance $100 \mu \mathrm{F}$ is charged at a steady rate of $100 \mu \mathrm{C} \mathrm{s}^{-1}$, then the time taken to produce a potential difference of $100 \mathrm{~V}$ between the capacitor plates is
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Verified Answer
The correct answer is:
$100 \mathrm{~s}$
Total charge on capacitor,
$$
\begin{aligned}
& Q=C V \\
& =(100 \mu \mathrm{F})(100 \mathrm{~V}) \\
& =100 \times 10^{-6} \times 100=10^{-2} \mathrm{C}
\end{aligned}
$$
Total charge $=Q($ Rate of charge $) \times($ time taken $)$
$$
\begin{aligned}
& \mathrm{Q}=\mathrm{Q}^{\prime} \times \mathrm{t} \\
& \mathrm{t}=\frac{\mathrm{Q}}{\mathrm{Q}^{\prime}}=\frac{10^{-2}}{100 \times 10^{-6}}=\frac{10^{-2}}{10^{-4}}=100 \\
& \mathrm{t}=100 \mathrm{sec}
\end{aligned}
$$
$$
\begin{aligned}
& Q=C V \\
& =(100 \mu \mathrm{F})(100 \mathrm{~V}) \\
& =100 \times 10^{-6} \times 100=10^{-2} \mathrm{C}
\end{aligned}
$$
Total charge $=Q($ Rate of charge $) \times($ time taken $)$
$$
\begin{aligned}
& \mathrm{Q}=\mathrm{Q}^{\prime} \times \mathrm{t} \\
& \mathrm{t}=\frac{\mathrm{Q}}{\mathrm{Q}^{\prime}}=\frac{10^{-2}}{100 \times 10^{-6}}=\frac{10^{-2}}{10^{-4}}=100 \\
& \mathrm{t}=100 \mathrm{sec}
\end{aligned}
$$
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