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

0.17

Step-by-step explanation:

correct answer on edge

Answer 2

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 = (\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 = ((8.85x10^(-12) )(28.27x10^(-6)))/(1.5x10^(-3) ) = 0.166pF`