Question

A parallel-plate capacitor has an area of 2.0 $c{m}^2$ and the plates are separated by 2.0 mm. What is the capacitance?

Answer 1

C=8.854×10−4μF

Explanation:

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

$C=\epsilon \frac{A}{d}$

where A is the area of each plate (the same for both), d is the distance which separates the two plates, and ε0 is a constant (vacuum permittivity) equal to

$8.854\times {10}^{-12}{C}^2/N{m}^2$

Given:

$A=2.0c{m}^{2}d=2mm$

$C=8.854\times {10}^{-12}{C}^2/N{m}^2.\frac{2.0\times {10}^{-4}}{2.0\times {10}^{-3}}=8.854\times {10}^{-10}F$

which is

$8.854\times {10}^{-4}\mu F$

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