When a $0.75 \mu \mathrm{F}$ capacitor is charged to a voltage of $20 \mathrm{~V}$, then the magnitude of…
When a $0.75 \mu \mathrm{F}$ capacitor is charged to a voltage of $20 \mathrm{~V}$, then the magnitude of charge on each plate is
$15 \mu \mathrm{C}$
$10 \mu \mathrm{C}$
$20 \mu \mathrm{C}$
$12 \mu \mathrm{C}$
Solution
Given that, capacitance of capacitor,
$
C=0.75 \mu \mathrm{F}
$
Voltage applied, $V=20 \mathrm{~V}$
Using an expression of charge stored on each plate of capacitor, $Q=C V$
Substituting the values, we get
$
\begin{aligned}
& Q=0.75 \times 10^{-6} \times 20=15 \times 10^{-6} \mathrm{C} \\
& Q=15 \mu \mathrm{C}
\end{aligned}
$
Hence, $15 \mu \mathrm{C}$ of charge stored on each plate of capacitor