Q163 / 170 · Manipal (MET) 2017 · Physics

Two capacitors, one 4 pF and the other 6 pF , connected in parallel, are charged by a 100 V battery. The energy stored in the capacitors is

A
$12 \times 10^{-8} \mathrm{~J}$
B
$2.4 \times 10^{-8} \mathrm{~J}$
Correct answer
C
$5.0 \times 10^{-8} \mathrm{~J}$
D
$1.2 \times 10^{-6} \mathrm{~J}$

Solution

The energy stored in capacitor is given by $E=\frac{1}{2} C V^2$ Resultant capacitance $\begin{aligned} C^{\prime} & =C_1+C_2=4+6=10 \mathrm{pF} \\ E & =\frac{1}{2} \times 10 \times 10^{-12} \times(100)^2 \left(1 \mathrm{pF}=10^{-12} \mathrm{~F}\right) \\ & =5 \times 10^{-8} \mathrm{~J} \end{aligned}$

Hence, the correct answer is option: 'C'.

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