Chapter 16 - Electric Potential, Energy, and Capacitance - Learning Path Questions and Exercises - Multiple Choice Questions - Page 590: 18

(a) it increases.

General equation connecting charge and potential difference is $q=CV$ C is constant for the capacitor (as $\kappa$,d and A are kept constant). Given that $V_{1}=6\,V$ and $V_{2}=12\,V$. $q_{1}=C\times6\,V$ $q_{2}=C\times12\,V$ Comparing $q_{1}$ and $q_{2}$, we find that $q_{2}\gt q_{1}$. That is, charge on the plate increases.