Respuesta :
Explanation:
Given that,
Diameter of the oil droplet, [tex]d=0.8\ \mu m[/tex]
Radius, [tex]r=0.4\ \mu m=0.4\times 10^{-6}\ m[/tex]
Separation between the electrodes, d = 11 mm
The droplet hangs motionless if the upper electrode is 22.8 V more positive than the lower electrode.
The density of the oil is [tex]885\ kg/m^3[/tex]
(a) Density of oil is given by mas per unit volume.
[tex]d=\dfrac{m}{V}\\\\m=d\times V\\\\m=d\times \dfrac{4}{3}\pi r^3\\\\m=885\times \dfrac{4}{3}\pi (0.4\times 10^{-6})^3\\\\m=2.37\times 10^{-16}\ kg[/tex]
(b) The electric force is balanced by the mass of the object. So,
qE = mg
[tex]q=\dfrac{mg}{E}[/tex]
E is electric field, E = V/d
[tex]q=\dfrac{mgd}{V}\\\\q=\dfrac{2.37\times 10^{-16}\times 9.8\times 11\times 10^{-3}}{22.8}\\\\q=1.12\times 10^{-18}\ C[/tex]
(c) Charges come in integer multiples of the electronic charge units of e. So,
[tex]n=\dfrac{q}{e}\\\\n=\dfrac{1.12\times 10^{-18}}{1.6\times 10^{-19}}\\\\n=7[/tex]
Hence, this is the required solution.
