Question

You invented an air capacitor. Its parallel plates are separated by 2.00 mm. Each plate carries a charge of 5.50 nC. The magnitude of the electric field of the plates is 5.50 x 10⁵ V/m. Find the capacitance of your invented air capacitor.

Answer 1

Given:

$d=0.002\:\rm m$

$q=5.50*10^{-9}\:\rm C$

$E=5.50*10^5\:\rm V/m$

The capacitance

$$C=\frac{q}{V}=\frac{q}{Ed}\\ =\frac{5.50*10^{-9}}{5.50*10^{5}*0.002}=5*10^{-12}\:\rm F$$