In figure, charge on the capacitor is plotted against potential difference across the capacitor. The capacitance and energy stored in the capacitor are respectively.

Question

Quantrex Academy · Accepted Answer

Correct answer: B. 12 μ F, 600 μ 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
In figure, charge on the capacitor is plotted against potential difference across the capacitor. The capacitance and

In figure, charge on the capacitor is plotted against potential difference across the capacitor. The capacitance and

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}

In figure, charge on the capacitor is plotted against potential...

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