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  
\[C=\frac{\epsilon_0 A}{d}\]
What happens to the voltage?
If the charge per unit surface area is  
  then the charge stored on the capacitor is  
\[Q= \sigma A\]

Substitute this last equation and (1) into the equation  

\[\sigma_0 A = V \frac{\epsilon_0 A}{d}\]

We can cancel  
  from both sides.
\[\sigma_0 = V \frac{\epsilon_0 }{d}\]

This equation is independent of  
  and reflects that increasing the area increases the capacitance and charge in proportion. typically the voltage is kept constant anyway by a batter.

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