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

Step-by-step 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

$C = \epsilon_0\dfrac{A}{d}$

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

$Q = \left(\epsilon_0\dfrac{A}{d}\right)V$

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

$\:\:\:\:\:=8.0×10^{-12}\:\text{C}$