The figure shows two identical parallel plate capacitors \(A\) and \(B\) of capacitances \(C\) connected to a battery. The key \(\mathrm{K}\) is initially closed. The switch is now opened and the free spaces between the plates of the capacitors are filled with a dielectric constant 3. Then which of the following statement (s) is/are true?

• Correct answer
  A

  Which the switch is closed, total energy stored in the two capacitors is \(\mathrm{CV}^2\)
• B

  When the switch is opened, no charge is stored in the capacitor B
• C

  When the switch is opened, energy stored in the capacitor \(\mathrm{B}\) is \(\frac{3}{2} \mathrm{CV}^2\)
• Correct answer
  D

  When the switch is opened, total energy stored in two capacitors is \(\frac{5}{2} \mathrm{CV}^2\)

Solution

Hint :
\(\begin{aligned} & \mathrm{U}=\frac{1}{2} \times 2 \mathrm{CV}^2 \\ & =\mathrm{CV}^2\end{aligned}\)
\(\begin{aligned} & \mathrm{U}_{\mathrm{A}}=\frac{1}{2}(\mathrm{KC}) \mathrm{V}^2=\frac{3}{2} \mathrm{CV}^2 \\ & \mathrm{U}_B=\frac{\mathrm{q}^2}{2 \mathrm{KC}}=\frac{[\mathrm{CV}]^2}{2 \mathrm{KC}}=\frac{\mathrm{CV}^2}{2 \mathrm{~K}}=\frac{1}{6} \mathrm{CV}^2\end{aligned}\) \(\underline{\text{Total energy when switch is open}}\) \(\begin{aligned} & \mathrm{U}=\frac{1}{2} \mathrm{KCV}^2+\frac{1}{6} \mathrm{CV}^2 \\ & =\frac{3}{2} \mathrm{CV}^2+\frac{1}{6} \mathrm{CV}^2 \\ & =\frac{10}{6} \mathrm{CV}^2=\frac{5}{3} \mathrm{CV}^2 \end{aligned}\)

Hence, the correct answer is options: 'A', 'D'.