CUET Preparation Today
| If the capacitor is charged in presence of dielectric and then the dielectric is removed afterwards, what will be the change in potential ? |
potential increases potential decreases remains constant can't say |
| potential increases |
$C = \frac{\epsilon_0 K A}{d}$ $V = \frac{Q}{C}$ $C_{\text{with dielectric}} = K C_0$ $V_{\text{initial}} = \frac{Q}{K C_0}$ $C_{\text{after removal}} = C_0$ $V_{\text{final}} = \frac{Q}{C_0}$ $V_{\text{final}} = K \cdot V_{\text{initial}}$ The potential increases. |
