Question

Oil has a density ρ of 860kg/m3. an oil droplet is suspended between two plates 1.0cm apart by adjusting the potential difference between them to 1177v. when the voltage is removed, the droplet falls and quickly reaches constant speed. it is timed with a stopwatch and falls 3.00mm in 7.33s. the viscosity of air is 1.83×10−5kg/m⋅s. what is the droplet's charge q?

Answer 1

After reaching the terminal velocity the speed of the oil drop is given as

$v = \frac{d}{t}$

it drop by 3 mm in 7.33 s

$v = \frac{0.003}{7.33}$

$v = 4.1 * 10^-4 m/s$

now by the formula of terminal speed

$6\pi \eta r v = \rho* \frac{4}{3}\pi r^3g$

$v = \frac{2}{9/eta}r^2\rho g$

$4.1*10^{-4} = \frac{2}{9*1.83 * 10^{-5}}*r^2*860*9.8$

$r = 0.16 mm$

so the radius of drop is 0.16 mm

Now the electric field between the plates is given as

$E = \frac{\Delta V}{\Delta x}$

$E = \frac{1177}{0.01} = 1.177* 10^5 N/C$

now we will use the condition of equilibrium

$qE = mg$

$q*1.177 * 10^5 = \rho* \frac{4}{3}\pi r^3 *9.8$

$q*1.177 * 10^5 = 860* \frac{4}{3}\pi *(0.16*10^{-3})^3*9.8$

$q = 1.2 * 10^{-12} C$

so above is the charge on the drop