Question

A parallel plate air capacitor of capacitance \( C \) is connected to a cell of emf \( V \) and then disconnected from it. A dielectric slab of dielectric constant \( K \), which can just fill the air gap of the capacitor, is now inserted in it. Which of the following is incorrect?

• A. The energy stored in the capacitor decreases \( K \) times.
• B. The change in energy stored is \( \frac{1}{2} C V^2 (1 - \frac{1}{K}) \).
• C. The charge on the capacitor is not conserved.
• D. The potential difference between the plates decreases \( K \) times.

Hint:

For isolated capacitors:
- Charge remains constant after disconnection.
- Energy decreases due to dielectric insertion.

Answer 1

The Correct Option is C

Step 1: Initial conditions.
The initial capacitance of the air capacitor is \( C \), and it is connected to a cell of emf \( V \). The initial charge on the capacitor is:
\[ Q_i = CV \] The initial energy stored in the capacitor is:
\[ U_i = \frac{1}{2} CV^2 \] Step 2: Effect of inserting a dielectric slab.
When a dielectric slab of dielectric constant \( K \) is inserted, the new capacitance \( C' \) becomes:
\[ C' = KC \] Since the capacitor is disconnected from the cell, the charge remains constant:
\[ Q_f = Q_i = CV \] Step 3: New potential difference.
The new potential difference \( V' \) across the capacitor is:
\[ V' = \frac{Q_f}{C'} = \frac{CV}{KC} = \frac{V}{K} \] Step 4: New energy stored in the capacitor.
The new energy stored in the capacitor is:
\[ U_f = \frac{1}{2} C' V'^2 = \frac{1}{2} KC \left( \frac{V}{K} \right)^2 = \frac{1}{2} KC \frac{V^2}{K^2} = \frac{1}{2} \frac{C V^2}{K} \] Step 5: Comparison of initial and final energies.
The initial energy stored was:
\[ U_i = \frac{1}{2} CV^2 \] The final energy stored is:
\[ U_f = \frac{1}{2} \frac{C V^2}{K} \] The energy decreases by a factor of \( K \):
\[ U_f = \frac{U_i}{K} \] Step 6: Change in energy stored.
The change in energy stored is:
\[ \Delta U = U_i - U_f = \frac{1}{2} CV^2 - \frac{1}{2} \frac{C V^2}{K} = \frac{1}{2} CV^2 \left(1 - \frac{1}{K}\right) \] Step 7: Incorrect statement analysis.
- (A) The energy stored decreases \( K \) times. This is correct.
- (B) The change in energy stored is \( \frac{1}{2} CV^2 \left(1 - \frac{1}{K}\right) \). This is correct.
- (C) The charge on the capacitor is not conserved. This is incorrect, as the charge remains constant.
- (D) The potential difference between the plates decreases \( K \) times. This is correct.
Thus, the incorrect statement is:
\[ (C) {The charge on the capacitor is not conserved.} \]