Question

Switch $ S $ is closed. Now the switch is opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant 3. The ratio of total electrostatic energy stored in the capacitors before and after the introduction of the dielectric is:

• A. \( 3 : 1 \)
• B. \( 5 : 1 \)
• C. \( 3 : 5 \)
• D. \( 5 : 3 \)

Hint:

When a dielectric is introduced after disconnecting the battery, the charge remains constant, and the energy changes due to the change in capacitance.

Answer 1

The Correct Option is A

We need to find the ratio of the total electrostatic energy stored in two capacitors before and after introducing a dielectric.
Step 1: Initial energy (switch closed). Capacitors \( A \) and \( B \) are in parallel, each with capacitance \( C \), connected to a battery of voltage \( V \). Total capacitance: \[ C_{\text{total, initial}} = C + C = 2C \] Initial energy: \[ U_{\text{initial}} = \frac{1}{2} (2C) V^2 = C V^2 \]
Step 2: After opening the switch. The switch is opened, so the total charge is conserved: \[ Q_{\text{total}} = 2C V \]
Step 3: After introducing the dielectric. Dielectric constant \( \kappa = 3 \). New capacitance of each capacitor: \[ C_{\text{new}} = 3C \] Total capacitance: \[ C_{\text{total, final}} = 3C + 3C = 6C \] New voltage: \[ V_{\text{new}} = \frac{2C V}{6C} = \frac{V}{3} \] Final energy: \[ U_{\text{final}} = \frac{1}{2} (6C) \left( \frac{V}{3} \right)^2 = \frac{1}{2} (6C) \cdot \frac{V^2}{9} = \frac{C V^2}{3} \] Step 4: Compute the ratio. \[ \frac{U_{\text{initial}}}{U_{\text{final}}} = \frac{C V^2}{\frac{C V^2}{3}} = 3 \implies U_{\text{initial}} : U_{\text{final}} = 3 : 1 \] Final Answer: \[ \boxed{3 : 1} \]