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A capacitor of $10 \mu \mathrm{F}$ is connected to a $10 \mathrm{~V}$ cell. The maximum charge on the capacitor will be
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Verified Answer
The correct answer is:
$100 \mu \mathrm{C}$
Given, $C=10 \mu \mathrm{F}=10 \times 10^{-6} \mathrm{~F}$
$V=10 \mathrm{~V}$
$\begin{aligned} \text{Change,} \quad q &=C \times V=10 \times 10^{-6} \times 10 \\ &=10^{-4} \mathrm{C} \\ &=100 \times 10^{6} \mathrm{C} \\ &=100 \mu \mathrm{C} \end{aligned}$
$V=10 \mathrm{~V}$
$\begin{aligned} \text{Change,} \quad q &=C \times V=10 \times 10^{-6} \times 10 \\ &=10^{-4} \mathrm{C} \\ &=100 \times 10^{6} \mathrm{C} \\ &=100 \mu \mathrm{C} \end{aligned}$
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