Answer to Question #349689 in Electric Circuits for Taxus

Question

A charged oil-drop of radius 1.3×10-⁶ m is prevented from falling under gravity by the vertical field between two horizontal plates charged to a difference of of 8340V. The distance between the plates is 16 mm and the density of oil is 920 kgm-³. Calculate the magnitude of the charge on the drop.

[Take g = 10 ms-²]

Answer 1

Given:

$\rho=920\frac{kg}{m^{3}}$ - density of the oil;

$r=1.3\times10^{-6}m$ - radius of the oil drop;

$g=10\frac{m}{s^{2}};$

$U=8340V;$

$d=16mm=16\times10^{-3}m$ - distance between plates;

Solution:

If electric field prevent falling oil drop, it means gravity force acting to oil drop equal to electric force. So,

$F_e=mg;$ $F_e=\frac{Uq}{d};$ here $q$ - charge of oil drop;

$m=\rho V=\rho(\frac{4}{3}\pi r^{3});$

$q=\frac{4\pi r^{3}\rho gd}{3U}=\frac{4\times3.14\times(1.3\times10^{-6}m)^{3}\times920\frac{kg}{m^{3}}\times10\frac{m}{s^{2}}\times16\times10^{-3}m}{3\times8340V}=162\times10^{-21}C;$

Answer:

$q=162\times10^{-21}C.$