Question #349689

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]

Expert's answer

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."

Learn more about our help with Assignments: Electric Circuits

## Comments

## Leave a comment