Question

4. MILLIKAN EXPERIMENT. An oil drop with a mass of 3.1 x10^-15kg absorbs 11 electrons, and then falls through an opening in two parallel, horizontal plates 1.8 cm apart. The upper plate is positive.a) Determine the charge absorbed by the oil drop. b) Determine the required potential difference between the plates in order for the oil drop to be suspendedbetween the two plates.

Answer 1

Given:

The mass of the oil drop is

$m\text{ = 3.1}\times10^{-15}\text{ kg}$

The number of electrons is n = 11

The distance between the plates is

$\begin{gathered} d=\text{ 1.8 cm} \\ =\text{ 0.018 m} \end{gathered}$

To find

(a) The charge absorbed by the oil drop.

(b) The potential difference between the plates.

Explanation:

The charge absorbed by the oil drop can be calculated by the formula

$Q=\text{ ne}$

Here, e is the charge of the electron whose magnitude will be

$e=\text{ 1.6}\times10^{-19}\text{ C}$

On substituting the values, the charge absorbed by the oil drop will be

$\begin{gathered} Q=11\times1.6\times10^{-19}\text{ } \\ =\text{ 1.76}\times10^{-18}\text{ C} \end{gathered}$

(b) The potential difference can be calculated as

$\begin{gathered} mg=\frac{QV}{d} \\ V=\frac{mgd}{Q} \\ =\frac{3.1\times10^{-15}\times9.8\times0.018}{1.76\times10^{-18}} \\ =310.704\text{ V} \end{gathered}$