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In figure, charge on the capacitor is plotted against potential difference across the capacitor. The capacitance and energy stored in the capacitor are respectively.
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Verified Answer
The correct answer is:
$12 \mu \mathrm{F}, 600 \mu \mathrm{J}$
Energy stored in a capacitor is given as
$U=\frac{1}{2} C V^{2}...(i)$
where, $C$ is the capacitance. From the given graph, we get
Slope $=\tan \theta=\frac{Q}{V}=\frac{120 \mu \mathrm{C}}{10 \mathrm{~V}}...(ii)$
As we know, capacitance is given as
$C=\frac{Q}{V}$
Substituting the value of $\frac{Q}{V}$ from Eq. (ii) in the above relation, we get
$\begin{aligned}
C &=\frac{120 \mu \mathrm{C}}{10 \mathrm{~V}}=\frac{120 \times 10^{-6}}{10}=12 \times 10^{-6} \mathrm{~F} \\
&=12 \mu \mathrm{F}
\end{aligned}$
Now, substituting the values of $C$ and $V$ in Eq. (i), we get
$\begin{aligned}
U &=\frac{1}{2} \times 12 \times(10)^{2} \\
&=600 \mu \mathrm{J}
\end{aligned}$
$U=\frac{1}{2} C V^{2}...(i)$
where, $C$ is the capacitance. From the given graph, we get
Slope $=\tan \theta=\frac{Q}{V}=\frac{120 \mu \mathrm{C}}{10 \mathrm{~V}}...(ii)$
As we know, capacitance is given as
$C=\frac{Q}{V}$
Substituting the value of $\frac{Q}{V}$ from Eq. (ii) in the above relation, we get
$\begin{aligned}
C &=\frac{120 \mu \mathrm{C}}{10 \mathrm{~V}}=\frac{120 \times 10^{-6}}{10}=12 \times 10^{-6} \mathrm{~F} \\
&=12 \mu \mathrm{F}
\end{aligned}$
Now, substituting the values of $C$ and $V$ in Eq. (i), we get
$\begin{aligned}
U &=\frac{1}{2} \times 12 \times(10)^{2} \\
&=600 \mu \mathrm{J}
\end{aligned}$
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