Chapter 4, Problem 6P

(a)

To determine

The radius and the mass of the drop used in the experiment.

Answer to Problem 6P

The radius is $\underset{\_}{1.62\times {10}^{-6}\text{m}}$ and the mass is $\underset{\_}{1.42\times {10}^{-14}\text{kg}}$.

Explanation of Solution

Write the expression for the radius of the oil drop in the Millikan’s experiment.

    $a=\sqrt{\frac{9\eta v}{2\rho g}}$        (I)

Here, $\eta$ is the viscosity of the liquid, $v$ is the velocity of the oil drop, $\rho$ is the density and $g$ is the acceleration due to gravity.

The mass of the oil drop is given by,

    $m=\rho V$        (II)

Here, $V$ is the volume, $m$ is the mass of the oil drop.

The volume of the oil drop is given by,

    $V=\frac{4}{3}\pi a^3$        (III)

Use equation (III) in (II),

    $m=\rho \left(\frac{4}{3}\right)\pi a^3$        (IV)

The average speed of the oil drop is given as,

    $\begin{array}{c}v=\frac{\Delta y}{\Delta t}\\ =\frac{4.0\times {10}^{-3}\text{m}}{15.9\text{s}}\\ =2.52\times {10}^{-4}\text{m}/\text{s}\end{array}$

Conclusion:

Substitute $1.81\times {10}^{-5}\text{kg}/\text{ms}$ for $\eta$, $2.52\times {10}^{-4}\text{m}/\text{s}$ for $v$, $800\text{kg}/{\text{m}}^3$ for $\rho$ and $9.81{\text{m}/\text{s}}^2$ for $g$ in equation (I) to find $a$.

    $\begin{array}{c}a={\left(\frac{9\left(1.81\times {10}^{-5}\text{kg}/\text{ms}\right)\left(2.52\times {10}^{-4}\text{m}/\text{s}\right)}{2\left(800\text{kg}/{\text{m}}^3\right)\left(9.81\text{m}/{\text{s}}^2\right)}\right)}^{1/2}\\ =1.62\times {10}^{-6}\text{m}\end{array}$

Substitute $1.33$ for $\rho$, $1.62\times {10}^{-6}\text{m}$ for $a$ in equation (IV) to find $m$.

    $\begin{array}{c}m=(1.33)\left(800\text{kg}/{\text{m}}^3\right)(\pi ){\left(1.62\times {10}^{-6}\text{m}\right)}^3\\ =1.42\times {10}^{-14}\text{kg}\end{array}$

Therefore, the radius is $\underset{\_}{1.62\times {10}^{-6}\text{m}}$ and the mass is $\underset{\_}{1.42\times {10}^{-14}\text{kg}}$.

(b)

To determine

The charge on each drop and show that the charge is quantized by considering both the size of each charge and the amount of charge gained when the rise time changes.

Answer to Problem 6P

It is shown that the charge is quantized by considering both the size of each charge and the amount of charge gained when the rise time changes.

Explanation of Solution

The charge on the droplet is given by,

    $q_1=\frac{\frac{mg}{E}\left(v+v_1{}^{\prime }\right)}{v}$        (V)

Here, $v$ is the velocity of fall, $v^{\prime }$ is the velocity of rise, $E$ is the electric field, $v^{\prime }$ is the upward terminal speed in the presence of the electric field, $v$ is the downward terminal velocity in the absence of the electric field.

The expression for the electric field is given by,

    $E=\frac{V}{d}$        (VI)

Conclusion:

Using the times given, the various $v^{\prime }$ can be calculated as,

    $\begin{array}{c}{v}_1{}^{\prime }=\frac{4\times {10}^{-3}\text{m}}{36.0\text{s}}\\ =1.11\times {10}^{-4}\text{m}/\text{s}\\ v_2{}^{\prime }=\frac{4\times {10}^{-3}\text{m}}{17.3\text{s}}\\ =2.31\times {10}^{-4}\text{m}/\text{s}\end{array}$

    $\begin{array}{c}{v}_3{}^{\prime }=\frac{4\times {10}^{-3}\text{m}}{24.0\text{s}}\\ =1.67\times {10}^{-4}\text{m}/\text{s}\\ v_4{}^{\prime }=\frac{4\times {10}^{-3}\text{m}}{11.4\text{s}}\\ =3.51\times {10}^{-4}\text{m}/\text{s}\end{array}$

    $\begin{array}{c}{v}_5{}^{\prime }=\frac{4\times {10}^{-3}\text{m}}{7.54\text{s}}\\ =5.31\times {10}^{-4}\text{m}/\text{s}\end{array}$

The corresponding charges can be written as,

    $\begin{array}{c}{q}_1=\frac{\left(6.97\times {10}^{-19}\text{C}\right)(2.52+1.11)}{2.52}\\ =10.0\times {10}^{-19}\text{C}\\ q_2=13.4\times {10}^{-19}\text{C}\end{array}$

    $\begin{array}{c}{q}_3=11.6\times {10}^{-19}\text{C}\\ q_4=16.7\times {10}^{-19}\text{C}\\ q_5=21.6\times {10}^{-19}\text{C}\end{array}$

Therefore, It is shown that the charge is quantized by considering both the size of each charge and the amount of charge gained when the rise time changes.

(c)

To determine

The electronic charge from the data.

Answer to Problem 6P

The average value is $\underset{\_}{1.67\times {10}^{-19}\text{C}}$.

Explanation of Solution

To find an integer $n_1$ such that $1.5\times {10}^{-19}<\frac{q}{n_1}<2.0\times {10}^{-19}$

    $\begin{array}{l}\frac{q_1}{6}=1.67\times {10}^{-19}\text{C}\\ \frac{q_2}{8}=1.68\times {10}^{-19}\text{C}\\ \frac{q_3}{7}=1.66\times {10}^{-19}\text{C}\\ \frac{q_4}{10}=1.67\times {10}^{-19}\text{C}\end{array}$

  $\frac{q_5}{13}=1.66\times {10}^{-19}\text{C}$

The amount of charge gained or lost is,

    $\begin{array}{l}{q}_2-q_1=3.4\times {10}^{-19}\text{C}\\ q_3-q_1=1.6\times {10}^{-19}\text{C}\\ q_2-q_3=1.8\times {10}^{-19}\text{C}\\ q_4-q_1=6.7\times {10}^{-19}\text{C}\end{array}$

    $\begin{array}{l}{q}_4-q_3=5.1\times {10}^{-19}\text{C}\\ q_4-q_2=3.4\times {10}^{-19}\text{C}\\ q_5-q_1=11.6\times {10}^{-19}\text{C}\\ q_5-q_4=4.9\times {10}^{-19}\text{C}\end{array}$

    $\begin{array}{l}{q}_5-q_4=4.9\times {10}^{-19}\text{C}\\ q_5-q_3=10.0\times {10}^{-19}\text{C}\end{array}$

The integers that yield a value of $e$ between 1.5 and $2.0\times {10}^{-19}\text{C}$.

    $\begin{array}{c}\frac{q_2-q_1}{2}=1.70\times {10}^{-19}\text{C}\\ \frac{q_3-q_1}{1}=1.60\times {10}^{-19}\text{C}\\ \frac{q_2-q_3}{1}=1.80\times {10}^{-19}\text{C}\\ \frac{q_4-q_1}{4}=1.68\times {10}^{-19}\text{C}\end{array}$

    $\begin{array}{c}\frac{q_4-q_3}{3}=1.70\times {10}^{-19}\text{C}\\ \frac{q_4-q_2}{2}=1.65\times {10}^{-19}\text{C}\\ \frac{q_5-q_1}{7}=1.66\times {10}^{-19}\text{C}\\ \frac{q_5-q_4}{3}=1.63\times {10}^{-19}\text{C}\end{array}$

    $\frac{q_5-q_3}{6}=1.67\times {10}^{-19}\text{C}$

The average of all values will be $1.67\times {10}^{-19}\text{C}$.

Conclusion:

Therefore, The average value is $\underset{\_}{1.67\times {10}^{-19}\text{C}}$.