A) [tex]2.4\cdot 10^{-16}kg[/tex]
The radius of the oil droplet is half of its diameter:
[tex]r=\frac{d}{2}=\frac{0.80 \mu m}{2}=0.40 \mu m = 0.4\cdot 10^{-6}m[/tex]
Assuming the droplet is spherical, its volume is given by
[tex]V=\frac{4}{3}\pi r^3 = \frac{4}{3}\pi (0.4\cdot 10^{-6} m)^3=2.68\cdot 10^{-19} m^3[/tex]
The density of the droplet is
[tex]\rho=885 kg/m^3[/tex]
Therefore, the mass of the droplet is equal to the product between volume and density:
[tex]m=\rho V=(885 kg/m^3)(2.68\cdot 10^{-19} m^3)=2.4\cdot 10^{-16}kg[/tex]
B) [tex]1.5\cdot 10^{-18}C[/tex]
The potential difference across the electrodes is
[tex]V=17.8 V[/tex]
and the distance between the plates is
[tex]d=11 mm=0.011 m[/tex]
So the electric field between the electrodes is
[tex]E=\frac{V}{d}=\frac{17.8 V}{0.011 m}=1618.2 V/m[/tex]
The droplet hangs motionless between the electrodes if the electric force on it is equal to the weight of the droplet:
[tex]qE=mg[/tex]
So, from this equation, we can find the charge of the droplet:
[tex]q=\frac{mg}{E}=\frac{(2.4\cdot 10^{-16}kg)(9.81 m/s^2)}{1618.2 V/m}=1.5\cdot 10^{-18}C[/tex]
C) Surplus of 9 electrons
The droplet is hanging near the upper electrode, which is positive: since unlike charges attract each other, the droplet must be negatively charged. So the real charge on the droplet is
[tex]q=-1.5\cdot 10^{-18}C[/tex]
we can think this charge has made of N excess electrons, so the net charge is given by
[tex]q=Ne[/tex]
where
[tex]e=-1.6\cdot 10^{-19}C[/tex] is the charge of each electron
Re-arranging the equation for N, we find:
[tex]N=\frac{q}{e}=\frac{-1.5\cdot 10^{-18}C}{-1.6\cdot 10^{-19}C}=9.4 \sim 9[/tex]
so, a surplus of 9 electrons.
