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  4. A 500 μ F capacitor is charged at a rate of 125 μ C / sec. The potential difference across the capacitor will be 10 V after an interval of

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Q. A $500 \mu F$ capacitor is charged at a rate of $125 \mu C / sec$. The potential difference across the capacitor will be $10\, V $ after an interval of

Electrostatic Potential and Capacitance

Solution:

$Q=C V$
$\Rightarrow Q=500 \times 10^{-6} \times 10=5000 \times 10^{-6} C$
The time interval required to charge the capacitor to $5000 \times 10^{-6} C$ will be equal to the one required for producing potential difference of $10 V$.
$T =\frac{\text { Total charge }}{\text { Rate of charging }}$
$=\frac{5000 \times 10^{-6}}{125 \times 10^{-6}}$
$=40 \,sec$