Question

As shown in the figure, a dielectric of constant K is placed between the plates of a parallel plate capacitor and is charged to a potential V using a battery. If the dielectric is pulled out after disconnecting the battery from the capacitor, the final potential difference across the plates of the capacitor is

• A. \( \left(1 + \frac{1}{K}\right) 2V \)
• B. \( 2KV \)
• C. \( \frac{2V}{\left(1 + \frac{1}{K}\right)} \)
• D. \( \frac{V}{2}\left(1 + \frac{1}{K}\right) \)

Hint:

When a battery is disconnected, Charge \( Q \) remains constant. When connected, Potential \( V \) remains constant. Always identify which parameter is conserved.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:

The capacitor configuration consists of two capacitors in series: one with the dielectric and one with air (since the dielectric is only on one side as shown in diagram - wait, let's analyze the diagram carefully). The diagram shows a dielectric slab of thickness \( d/2 \) (implied by the arrows) between plates separated by \( d \). The battery charges it to \( V \), then is disconnected. Charge \( Q \) remains constant. When the dielectric is removed, the capacitance changes, changing the potential. However, looking at the options and the standard nature of such problems, the question implies a comparison between the initial state (with dielectric partially filling) and the final state (dielectric removed). Let's re-read carefully: "dielectric ... is placed ... charged ... dielectric is pulled out". Initial State: Composite capacitor. Final State: Air capacitor. Charge \( Q \) is constant. \( V_{\text{final}} = \frac{Q}{C_{\text{final}}} \).
Step 2: Key Formula or Approach:

1. Capacitance with partial dielectric (thickness \( t=d/2 \)): \( C_{\text{initial}} = \frac{\epsilon_o A}{d - t + t/K} = \frac{\epsilon_o A}{d/2 + d/(2K)} = \frac{2\epsilon_o A}{d(1 + 1/K)} \). 2. Charge \( Q = C_{\text{initial}} V \). 3. Final Capacitance (air): \( C_{\text{final}} = \frac{\epsilon_o A}{d} \). 4. Final Potential: \( V' = \frac{Q}{C_{\text{final}}} \).
Step 3: Detailed Explanation:

Let \( C_o = \frac{\epsilon_o A}{d} \). The initial capacitance \( C_i \) with dielectric of width \( d/2 \) (assumed from figure symmetry) is: \[ C_i = \frac{\epsilon_o A}{(d - d/2) + \frac{d/2}{K}} = \frac{\epsilon_o A}{\frac{d}{2} \left(1 + \frac{1}{K}\right)} = \frac{2\epsilon_o A}{d \left(1 + \frac{1}{K}\right)} = \frac{2C_o}{1 + \frac{1}{K}} \] Charge stored \( Q = C_i V \). When battery is disconnected, \( Q \) is constant. When dielectric is removed, the capacitor becomes a standard air capacitor with capacitance \( C_f = C_o \). New Potential \( V' = \frac{Q}{C_f} = \frac{C_i V}{C_o} \). Substitute \( C_i \): \[ V' = \frac{\frac{2C_o}{1 + 1/K} V}{C_o} = \frac{2V}{1 + \frac{1}{K}} \]
Step 4: Final Answer:

The final potential difference is \( \frac{2V}{1 + \frac{1}{K}} \).