Question

Two 4.0-cm-diameter aluminum electrodes are spaced 0.50 mm apart. the electrodes are connected to a 100 v battery. part a what is the capacitance?

Answer 1

The capacitance of a parallel-plate capacitor in vacuum is given by
$C= \frac{A \epsilon_0}{d}$
where
A is the area of each plate
d is the separation between the plates
$\epsilon_0 = 8.85 \cdot 10^{-12} F/m$ is the electric permittivity in vacuum

For the capacitor in our problem, the plates are circular, with a diameter of d=4.0 cm, so their radius is r=2.0 cm=0.02 m and their area is
$A=\pi r^2 =\pi (0.02 m)^2 = 1.26 \cdot 10^{-3} m^2$
while the separation between the plates is
$d=0.50 mm=5 \cdot 10^{-4} m$

Therefore the capacitance is
$C= \frac{A\epsilon_0 }{d}= \frac{(1.26 \cdot 10^{-3} m^2)(8.85 \cdot 10^{-12} F/m)}{5 \cdot 10^{-4}m}=2.23 \cdot 10^{-11}F$