Two identical capacitors A and B are connected in series to a battery of E.M.F., 'E'. Capacitor B contains a…

Two identical capacitors A and B are connected in series to a battery of E.M.F., 'E'. Capacitor B contains a slab of dielectric constant $\mathrm{K} . \mathrm{Q}_{\mathrm{A}}$ and $\mathrm{Q}_{\mathrm{B}}$ are the charges stored in A and B . When the dielectric slab is removed, the corresponding charges are $\mathrm{Q}_{\mathrm{A}}^{\prime}$ and $\mathrm{Q}_{\mathrm{B}}$. Then
  1. $\frac{Q_A^{\prime}}{Q_A}=\frac{K}{2}$
  2. $\frac{\mathrm{Q}_{\mathrm{B}}^{\prime}}{\mathrm{Q}_{\mathrm{B}}}=\frac{\mathrm{K}+1}{2}$
  3. $\frac{Q_A^{\prime}}{Q_A}=\frac{K+1}{K}$
  4. $\quad \frac{\mathrm{Q}_{\mathrm{B}}^{\prime}}{\mathrm{Q}_{\mathrm{B}}}=\frac{\mathrm{K}+1}{2 \mathrm{~K}}$

Solution

$\begin{aligned} & C_A=C, C_B=K C \\ \therefore \quad & C_{n e t}=\frac{C_A C_B}{C_A+C_B}=\frac{C \times K C}{C+K C}=\left(\frac{K}{K+1}\right) C \\ & Q_A=Q_B=\frac{K C V}{K+1} \end{aligned}$
After the slab is removed, $C_A=C ; C_B=C$ $\begin{aligned} \therefore \quad & \mathrm{C}_{\mathrm{net}}=\frac{\mathrm{C} \times \mathrm{C}}{\mathrm{C}+\mathrm{C}}=\frac{\mathrm{C}}{2} \\ & \mathrm{Q}_{\mathrm{A}}^{\prime}=\mathrm{Q}_{\mathrm{B}}^{\prime}=\frac{\mathrm{CV}}{2} \end{aligned}$ $\therefore \quad \frac{\mathrm{Q}_{\mathrm{B}}^{\prime}}{\mathrm{Q}_{\mathrm{B}}}=\frac{\frac{\mathrm{CV}}{2}}{\frac{\mathrm{KCV}}{\mathrm{~K}+1}}=\frac{\mathrm{K}+1}{2 \mathrm{~K}}$

Asked in: MHT CET 2024 (16 May Shift 1)

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