A battery is connected across a capacitor of 20 µF through a switch which is initially open.The 240 µC.of charge flows from the battery to the capacitor when the switch is closed.The electric potential provided by the battery is

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Quizplus · Accepted Answer

The electric potential (voltage) provided by the battery can be calculated using the formula $V = \frac{Q}{C}$, where $V$ is the voltage, $Q$ is the charge in coulombs, and $C$ is the capacitance in farads. Substituting the given values, $Q = 240 \mu C = 240 \times 10^{-6} C$ and $C = 20 \mu F = 20 \times 10^{-6} F$, we get $V = \frac{240 \times 10^{-6}}{20 \times 10^{-6}} = 12 V$.

A Battery Is Connected Across a Capacitor of 20 µF