Solution:

The solution can be derived as follows:

1. Initial Capacitance (with air as the dielectric):
   For a parallel plate capacitor with air as the dielectric, the capacitance $C_{\text{initial}}$ is given by:

   $$C_{\text{initial}} = C$$The battery applies a potential difference $E$, so the initial charge on the capacitor is:

   $$Q_{\text{initial}} = C_{\text{initial}} \times E = C \times E$$

2. Capacitance with Dielectric Slab:
   When a dielectric slab of dielectric constant $K$ is inserted between the plates, the capacitance increases by a factor of $K$, so the new capacitance $C_{\text{new}}$ becomes:

   $$C_{\text{new}} = K \times C$$

3. Final Charge on Capacitor:
   The battery remains connected and maintains a constant voltage $E$. Therefore, the final charge on the capacitor is:

   $$Q_{\text{final}} = C_{\text{new}} \times E = K \times C \times E$$

4. Additional Charge Drawn from the Battery:
   The additional charge drawn from the battery is the difference between the final charge and the initial charge:

   $$\Delta Q = Q_{\text{final}} - Q_{\text{initial}} = (K \times C \times E) - (C \times E)$$Simplifying:

   $$\Delta Q = (K - 1) \times C \times E$$

Thus, the additional charge drawn from the battery is:

$$(K - 1) \times C \times E$$

Discuss this question