Answer to Question #138562 in Electric Circuits for Katy

Question

A charged oil-drop of radius 1.3 x 10-4cm is prevented from falling under gravity by the vertical field between two horizontal plates charged to a difference of potential of 8340 volts. The distance between the plates is 1.6 cm, and the density of oil is 920kgm-3. Calculate the magnitude of the charge on the drop.

Answer 1

Volume of oil:

$V=\frac{4}{3}\times \pi \times R^3$

$V=\frac{4}{3}\times 3.14 \times (1.3\times10^{-6})^3=\frac{4}{3}\times 3.14 \times 2.197\times10^{-18}$

$V=9.2\times 10^{-18}m^3$

Mass of oil:

$\rho=\frac{m}{V}$

$m=\rho \times V=920\times 9.2 \times 10^{-18}=8.464 \times 10^{-15}$

Gravity force for oil:

$F_{g}=m\times g = 8.464 \times 10^ {-15}\times9.8 = 8.3 \times10^{-14}$

Electric field between plates:

$E=\frac{U}{d}=\frac{8340}{0.016}=521250$

Charge of oil:

$q=\frac{F}{E}=\frac{8.3\times10^{-14}}{521250}=1.6\times10^{-19}$