\[C=\frac{\epsilon_0 A}{d}\]
(1)What happens to the voltage?
If the charge per unit surface area is
\[\sigma\]
then the charge stored on the capacitor is \[Q= \sigma A\]
Substitute this last equation and (1) into the equation
\[Q=VC\]
\[\sigma_0 A = V \frac{\epsilon_0 A}{d}\]
We can cancel
\[A\]
from both sides.\[\sigma_0 = V \frac{\epsilon_0 }{d}\]
This equation is independent of
\[A\]
and reflects that increasing the area increases the capacitance and charge in proportion. typically the voltage is kept constant anyway by a batter.