Question

A parallel plate capacitor has an area of ​​1.5 cm

2
and the plates are separated a distance of 2.0 mm with air between them. How much charge does this capacitor store when connected to a 12V battery?

Answer 1

Explanation:

Given:

`A=1.5\:\text{cm}^2×\left(\frac{1\:\text{m}^2}{10^4\:\text{cm}^2}\right)=1.5×10^(-4)\:\text{m}^2`

`d = 2.0\:\text{mm} = 2.0×10^(-3)\:\text{mm}`

The charge stored in a capacitor is given by
`Q = CV.` In the case of a parallel-plate capacitor, its capacitance C is given by

`C = \epsilon_0(A)/(d)`

where
`\epsilon_0` = permittivity of free space. The amount of charge stored in the capacitor is then

`Q = \left(\epsilon_0(A)/(d)\right)V`

`\:\:\:\:\:=\left[\frac{(8.85×10^(-12)\:\text{F/m})(1.5×10^(-4)\:\text{m}^2)}{(2.0×10^(-3)\:\text{m})}\right](12\:\text{V})`

`\:\:\:\:\:=8.0×10^(-12)\:\text{C}`