Question

In 1909, Robert Millikan was the first to find the charge of an electron in his now-famous oil-drop experiment. In that experiment, tiny oil drops were sprayed into a uniform electric field between a horizontal pair of oppositely charged plates. The drops were observed with a magnifying eyepiece, and the electric field was adjusted so that the upward force on some negatively charged oil drops was just sufficient to balance the downward force of gravity. That is, when suspended, upward force just equaled . Millikan accurately measured the charges on many oil drops and found the values to be whole number multiples of the charge of the electron. For this, he won the Nobel Prize. Extra electrons on this particular oil drop (given the presently known charge of the electron) are:
A. 4
B. 3
C. 5
D. 8

Answer 1

Equate the electrostatic force acting on a drop and gravitational force acting on it to find the given ratio.

Complete step by step solution:
Millikan oil-drop experiment was performed by the American physicist Robert A. Millikan in 1909. He devised a straightforward method of measuring the electric charge present on many of the droplets in an oil mist. The force on any electric charge in an electric field is equal to the product of the charge and the electric field. Millikan was able to measure both the amount of electric force and magnitude of electric field on the tiny charge of an isolated oil droplet. In the Millikan oil-drop experiment a closed chamber with transparent sides is fitted with two parallel metal plates, which acquire a positive or negative charge when a voltage is applied across the metal plates.
It is given that the electrostatic force and the gravitational force acting on a drop balance each other.

Electrostatic force acting on the drop = $qE$
Gravitational force acting on the drop = $mg$
$\Rightarrow qE = mg$
$\Rightarrow q \times 1.68 \times {10^5} = 1.08 \times {10^{ - 14}}$
$\Rightarrow q = 6.4 \times {10^{ - 19}}$
Let n be the number of electrons in the drop. Then,
$ne = 6.4 \times {10^{ - 19}}$
$n = \dfrac{{6.4 \times {{10}^{ - 19}}}}{{1.6 \times {{10}^{ - 19}}}} = 4$

Hence, the correct option is (A).

Note: Millikan’s oil drop experiment measured the charge of an electron. Before this experiment, the existence of subatomic particles was not universally accepted. In the experiment, if we adjust the potential difference or voltage across the metal plates, the speed of the droplet’s motion can be increased or decreased. When the amount of upward electric force equals the downward gravitational force, the charged droplet remains stationary. The amount of voltage needed to suspend a droplet is used along with its mass to determine the overall electric charge on the droplet.