Question

The plates on a vacuum capacitor have a radius of 3.0 mm and are separated by a distance of 1.5 mm. What is the capacitance of this capacitor? Recall that ε0 = 8.85 × 10–12

Answer 1

Capacitance is the ratio of the change in an electric charge in a system to the corresponding change in its electric potential. For a Parallel-plate capacitor, the formula to calculate the capacitance is given by:

$c = \frac{\epsilon_{0}A}{d}$

being:

$\epsilon_{0}$: the electric constant 
$A$: the area of overlap of the two plates, in square meters
$d$:  he separation between the plates, in meters

Given that the plates of the capacitor have Circular Cross-Section, then:

$A = \pi r^{2} = \pi (3x10^{-3})^{2}=28.27x10^{-6}m^{2}$

Therefore, the capacitance is:

$c = \frac{(8.85x10^{-12} )(28.27x10^{-6})}{1.5x10^{-3} } = 0.166pF$

Answer 2

Answer:

0.17

Explanation:

correct answer on edge