If the capacitor is charged in presence of dielectric and then the dielectric is removed afterwards, what will be the change in potential ?
Answer & explanation
Correct answer: option 1
$C = \frac{\epsilon_0 K A}{d}$
$V = \frac{Q}{C}$
$C_{\text{with dielectric}} = K C_0$
$V_{\text{initial}} = \frac{Q}{K C_0}$
$C_{\text{after removal}} = C_0$
$V_{\text{final}} = \frac{Q}{C_0}$
$V_{\text{final}} = K \cdot V_{\text{initial}}$
The potential increases.