Question

Two parallel plate capacitors of capacitances \( C \) and \( 2C \) are joined with a battery of voltage difference \( V \) as shown in the figure. If the battery is removed and the space between the plates of the capacitor of capacitance \( C \) is completely filled with a material of dielectric constant \( K \), then find out:

• i) Total capacitance of the combination
• ii) Final voltage difference across the combination
• iii) Total energy stored in the combination.

Hint:

In capacitors, the total charge remains constant when a battery is removed, and energy calculations depend on the final capacitance and voltage configuration.

Answer 1

1. Total Capacitance of the Combination:

   The initial total capacitance (without dielectric) is the series combination of \( C \) and \( 2C \):

   \[ \frac{1}{C_{\text{total, initial}}} = \frac{1}{C} + \frac{1}{2C} = \frac{3}{2C} \quad \Rightarrow \quad C_{\text{total, initial}} = \frac{2C}{3}. \]

   After inserting the dielectric of constant \( K \) into the capacitor of capacitance \( C \), its new capacitance becomes:

   \[ C' = KC. \]

   The new total capacitance is the series combination of \( C' = KC \) and \( 2C \):

   \[ \frac{1}{C_{\text{total, final}}} = \frac{1}{KC} + \frac{1}{2C}. \]

   Simplifying:

   \[ \frac{1}{C_{\text{total, final}}} = \frac{K + 2}{2KC}. \]

   Thus:

   \[ C_{\text{total, final}} = \frac{2KC}{K + 2}. \]

2. Final Voltage Difference Across the Combination:

   The charge on the initial combination is given by:

   \[ Q_{\text{initial}} = C_{\text{total, initial}} \cdot V = \frac{2C}{3} \cdot V. \]

   Since the battery is removed, the total charge remains the same. The final voltage \( V_{\text{final}} \) across the new combination is:

   \[ V_{\text{final}} = \frac{Q_{\text{initial}}}{C_{\text{total, final}}}. \]

   Substituting the values:

   \[ V_{\text{final}} = \frac{\frac{2C}{3} \cdot V}{\frac{2KC}{K + 2}} = \frac{V \cdot (K + 2)}{3K}. \]

3. Total Energy Stored in the Combination:

   The total energy stored in a capacitor is given by:

   \[ U = \frac{1}{2} C_{\text{total, final}} \cdot V_{\text{final}}^2. \]

   Substituting \( C_{\text{total, final}} = \frac{2KC}{K + 2} \) and \( V_{\text{final}} = \frac{V \cdot (K + 2)}{3K} \):

   \[ U = \frac{1}{2} \cdot \frac{2KC}{K + 2} \cdot \left(\frac{V \cdot (K + 2)}{3K}\right)^2. \]

   Simplifying:

   \[ U = \frac{1}{2} \cdot \frac{2KC}{K + 2} \cdot \frac{V^2 \cdot (K + 2)^2}{9K^2}. \]

   \[ U = \frac{C V^2 \cdot (K + 2)}{9K}. \]

   Thus, the total energy stored is:

   \[ U = \frac{C V^2 (K + 2)}{9K}. \]