A system of \( 2 \) capacitors of capacitance \( 2 \mu \mathrm{F} \) and \( 4 \mu \mathrm{F} \) is connected in series across a potential
difference of \( 6 \mathrm{~V} \). The electric charge and energy stored in the system are

• A

  \( 10 \mu \mathrm{C} \) and \( 30 \mu \mathrm{J} \)
• B

  \( 36 \mu \mathrm{C} \) and \( 108 \mu \mathrm{J} \)
• Correct answer
  C

  \( 8 \mu \mathrm{C} \) and \( 24 \mu \mathrm{J} \)
• D

  \( 1 \mu C \) and \( 3 \mu J \)

Solution

The effective capacitance of system of two given capacitors is
\[ \begin{array}{l} =\frac{1}{2 \mu F}+\frac{1}{4 \mu F}=\frac{(2+1)}{4 \mu F}=\frac{3}{4} \\ \Rightarrow C=\frac{4}{3} \mu F \end{array} \]
Now, using \( Q=C V \)
\[ Q=\frac{4}{3} \times 10^{-6} \mathrm{~F} \times 6 \mathrm{~V}=8 \times 10^{-6} \mathrm{C} \]

Therefore, electric charge \( =8 \mu \mathrm{C} \)
Now, energy stored \( =\frac{1}{2} C V^{2}=\frac{1}{2} \times \frac{4}{3} \times 10^{-6} \times 6 \times 6 \)
\( =2 \times 10^{-6} \times 2 \times 6=24 \times 10^{-6} \mathrm{~J} \)
Therefore, energy stored \( =24 \mu \mathrm{J} \)

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