Question

In 1909 Robert Millikan was the first to find the charge of an electron in his now-famous oil drop experiment. In the experiment tiny oil drops are sprayed into a uniform electric field between a horizontal pair of oppositely charged plates. The drops are observed with a magnifying eyepiece, and the electric field is adjusted so that the upward force q E on some negatively charged oil drops is just sufficient to balance the downward force m g of gravity. Millikan accurately measured the charges on many oil drops and found the values to be whole-number multiples of 1.6 × 10−19 C — the charge of the electron. For this he won the Nobel Prize. If a drop of mass 1.51837 × 10−12 kg remains stationary in an electric field of 1 × 106 N/C, what is the charge on this drop? The acceleration due to gravity is 9.8 m/s 2 . Answer in units of C.

Answer 1

Answer:

$1.49\cdot 10^{-17}C$

Explanation:

The oil drop remains stationary when the electric force on it and the gravitational force are balanced, so we have:

$F_E = F_G\\qE = mg$

where

q is the charge of the oil drop

E is the electric field strength

m is the mass of the drop

g is the acceleration due to gravity

here we have

$E=1\cdot 10^6 N/C$

$m=1.51837\cdot 10^{-12} kg$

$g=9.8 m/s^2$

So the charge of the drop is

$q=\frac{mg}{E}=\frac{(1.51837\cdot 10^{-12} kg)(9.8 m/s^2)}{1\cdot 10^6 N/C}=1.49\cdot 10^{-17}C$