Answer to Question #181912 in Mechanics | Relativity for Jay

Question

A charged oil droplet of radius r = 1.5•10^-6 m is prevented from falling under gravity by the vertical electric field between two large horizontal plates maintained at a separation of 20mm and a potential difference of 100 V. The density of oil is 920kg/m³.

a) Find the charge on the oil droplet (g = 9.81m/s²).

b) If the polarity of the plates is reversed what is the instantaneous acceleration of the droplet?

Answer 1

Charge

$m= \rho V=\rho\times \frac{4 \pi}{3}r^3= 920\times \frac{4 \pi}{3}(1.5 \times10^{-6})^3$

$qE=mg; E =\frac{V}{d}=\frac{100}{20 \times10^{-3}}=5000$

$q= \frac{mg}{E}= 920 \times\frac{4 \pi}{3} \times\frac{1.5^3 \times 10^{-18} \times 9.8}{5000}$

$q=2.55 \times 10^{-17} C$

Instantaneous acceleration

$F=qE+mg =mg +mg=2mg$

$ma=2mg$

$a=2g$

$a=2 \times9.81 =19.62 m/s^2$