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 5.2898 × 10−13 kg remains stationary in an electric field of 6 × 105 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:

$8.64\cdot 10^{-18} C$

Explanation:

There are two forces acting on the oil drop:

- The force of gravity, downward, given by

$F_G = mg$

where m is the mass of the drop and g is the acceleration due to gravity

- The electric force, upward, given by

$F_E = qE$

where q is the charge of the oil drop and E is the magnitude of the electric field

The oil drop remains stationary, so the two forces are balanced:

$F_G = F_E\\mg = qE$

where

$m=5.2898\cdot 10^{-13}kg\\E=6\cdot 10^5 N/C\\g = 9.8 m/s^2$

Substituting into the previous equation and solving for q, we find the charge of the oil drop:

$q=\frac{mg}{E}=\frac{(5.2898\cdot 10^{-13} kg)(9.8 m/s^2)}{6\cdot 10^5 N/C}=8.64\cdot 10^{-18} C$