Chapter 27, Problem 3P

To determine

Number of excess electrons does the drop have?

Answer to Problem 3P

Solution:

The number of excess electron on the oil drop is $5$.

Explanation of Solution

Given:

Mass of the oil drop, $m=2.8\times {10}^{-15}\text{kg}$

Distance between the plates, $d=1\text{cm}=\text{0}\text{.01}\text{m}$

Potential difference between the plates, $V=340\text{volt}$

Formula used:

The weight of the oil drop is given as

$F=mg$

Here,

$m$ is the mass

$g$ is the acceleration due to gravity

The electric force on the drop is calculated as

$F=qE$

Here,

$q$ is the charge of the oil

$E$ is the electric field

Number of excess electron on the oil drop is calculated as

$q=ne$

Here,

$n$ is the number of excess electron

$e$ is the charge of the electron

Calculation:

The electric field is calculated as

$E=\frac{V}{d}$

Plugging the values in the above equation

$\begin{array}{c}E=\frac{340}{0.01}\\ =34000\text{V/m}\end{array}$

Since, the oil drop is held at rest; therefore, the electric force and force due to gravity on the drop must be equal

$qE=mg$

Plugging the values in the above equation

$\begin{array}{c}q\times 34000=2.8\times {10}^{-15}\times 9.81\\ q=8.0\times {10}^{-19}\text{C}\end{array}$

Number of excess electron on the oil drop is calculated as

$q=ne$

Plugging the values in the above equation

$\begin{array}{c}8\times {10}^{-19}=n\times 1.6\times {10}^{-19}\\ n=5\end{array}$