## 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.

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