## Effect Of Increasing Area on a Parallel Plate Capacitor

When the surface area of a capacitor increase, it's capacity to store charge (it's capacitance) increases according to the equation\[C=\frac{\epsilon_0 A}{d}\]

(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.