Question: Q. 11. A $12 p \mathrm{~F}$ capacitor is connected to a $50 \mathrm{~V}$ battery. How much electrostatic energy is stored in the capacitor? If another capacitor of $6 p \mathrm{~F}$ is connected in series with it with the same battery connected across the combination, find the charge stored and potential difference across each capacitor.
U [Delhi II 2017]
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Solution:
Ans. Calculation of electrostatic energy in $12 p \mathrm{~F}$ capacitor
Total charge stored in combination $\mathbf{1}$
Potential difference across each capacitor $1 / 2+1 / 2$
Energy stored, in the capacitor of capacitance 12 $p \mathrm{~F}$,
$$ \begin{aligned} U & =\frac{1}{2} C V^{2} \ & =\frac{1}{2} \times 12 \times 10^{-12} \times 50 \times 50 \mathrm{~J} \ & =1.5 \times 10^{-8} \mathrm{~J} \end{aligned} $$
$C=$ Equivalént capacitance of $12 p F$ and $6 p F$, in series, is given by
$$ \begin{aligned} & \frac{1}{C}=\frac{1}{12}+\frac{1}{6}=\frac{1+2}{12} \ & C=4 p F \end{aligned} $$
Charge stored across each capacitor
$$ \begin{aligned} q & =C V \ & =4 \times 10^{-12} \times 50 C \ & =2 \times 10^{-10} \mathrm{C} \end{aligned} $$
Charge on each capacitor $12 p \mathrm{~F}$ as well as $6 p \mathrm{~F}$
$\therefore$ Potential difference across capacitor $C_{1}$
$$ \begin{aligned} V_{1} & =\frac{2 \times 10^{-10}}{12 \times 10^{-12}} \ & =\frac{50}{3} \mathrm{~V} \end{aligned} $$
$\therefore$ Potential difference across capacitor $C_{2}$
$$ \begin{aligned} & \therefore \quad V_{2}=\frac{2 \times 10^{-10}}{6 \times 10^{-12}} \ & =\frac{100}{3} \mathrm{~V} \end{aligned} $$
[CBSE Marking Scheme 2017]
