Question

A parallel-plate capacitor has an area of `2.0` `cm^2` and the plates are separated by `2.0` `mm`. What is the capacitance?

Answer 1

`C=8.854xx10^-4 muF`

Explanation:

For a parallel plate capacitor, capacitance may be expressed by the following equation:

`C=epsilon_0A/d`

where `A` is the area of each plate (the same for both), `d` is the distance which separates the two plates, and `epsilon_0` is a constant (vacuum permittivity) equal to `8.854xx10^(-12)C^2//Nm^2`.

We are given that `A=2.0cm^2=2.0xx10^-4m^2` and `d=2.0mm=2.0xx10^-3m`. Therefore, we have:

`C=8.854xx10^(-12)C^2//Nm^2*(2.0xx10^-4m^2)/(2.0xx10^-3m)`

`=8.854xx10^-10F`

Which is `8.854xx10^-4 muF`

Farads, the unit of capacitance, is extremely large, so it is not uncommon to see the capacitance of a device expressed in microfarads.