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Consider an oil droplet of mass m and charge q. We want to determine the charge on the droplet in a Millikan-type experiment. We will do this in several steps. Assume, for simplicity, that the charge is positive and that the electric field between the plates points upward.

a. An electric field is established by applying a potential difference to the plates. It is found that a field of strength ${\mathrm{E}}_0$will cause the droplet to be suspended motionless. Write an expression for the droplet鈥檚 charge in terms of the suspending field ${\mathrm{E}}_0$and the droplet鈥檚 weight mg.

b. The field ${\mathrm{E}}_0$is easily determined by knowing the plate spacing and measuring the potential difference applied to them. The larger problem is to determine the mass of a microscopic droplet. Consider a mass m falling through a viscous medium in which there is a retarding or drag force. For very small particles, the retarding force is given by ${\mathrm{F}}_{\mathrm{drag}}=鈭�\mathrm{bv}$where b is a constant and v the droplet鈥檚 velocity. The sign recognizes that the drag force vector points upward when the droplet is falling (negative v). A falling droplet quickly reaches a constant speed, called the terminal speed. Write an expression for the terminal speed in ${\mathrm{v}}_{\mathrm{term}}$terms of m, g, and b.

c. A spherical object of radius r moving slowly through the air is known to experience a retarding force ${\mathrm{F}}_{\mathrm{drag}}=鈭�6\mathrm{蟺畏谤痴}$where $\mathrm{畏}$is the viscosity of the air. Use this and your answer to part b to show that a spherical droplet of density r falling with a terminal velocity ${\mathrm{v}}_{\mathrm{term}}$has a radius $\mathrm{r}=\sqrt{\frac{9{\mathrm{畏惫}}_{\mathrm{term}}}{2\mathrm{蚁驳}}}$

d. Oil has a density $860\mathrm{kg}/{\mathrm{m}}^3$. An oil droplet is suspended between two plates 1.0 cm apart by adjusting the potential difference between them to 1177 V. When the voltage is removed, the droplet falls and quickly reaches constant speed. It is timed with a stopwatch, and falls 3.00 mm in 7.33 s. The viscosity of air is . What is the droplet鈥檚 charge?

e. How many units of the fundamental electric charge does this droplet possess?

Step by step solution

01

Part (a) step 1: Given information

The droplet鈥檚 weight is mg

02

Part (a) Step 2: Calculations

On a positive charge, we can start with Newton's second law.:

$$\begin{gathered} \mathrm{m}\stackrel{鈫�}{\mathrm{a}}鈥呪赌呪赌呪赌�={\stackrel{鈫�}{\mathrm{F}}}_{\mathrm{E}}+{\stackrel{鈫�}{\mathrm{F}}}_{\mathrm{g}} \\ \mathrm{ma}鈥呪赌呪赌呪赌�={\mathrm{qE}}_0鈭�\mathrm{mg}鈥呪赌呪赌呪赌�\text{(expressions for electric and gravitational force)} \\ \mathrm{a}鈥呪赌呪赌呪赌�=0鈥呪赌呪赌呪赌�\left(\mathrm{chargesuspendedmotionless}\right) \\ {\mathrm{qE}}_0鈥呪赌呪赌呪赌�=\mathrm{mg}鈥呪赌呪赌呪赌�\text{(substitue}\mathrm{a}=0) \\ \mathrm{q}鈥呪赌呪赌呪赌�=\frac{\mathrm{mg}}{{\mathrm{E}}_0}\text{(express}\mathrm{q}\text{)} \end{gathered}$$

03

Part (b) Step 1: Given information

The retarding force is given by${\mathrm{F}}_{\mathrm{drag}}=鈭�\mathrm{bv}$

04

Part (b) Step 2: Calculations

With drag force, we can now construct Newton's second law:

$$\begin{gathered} {\stackrel{鈫�}{\mathrm{F}}}_{\mathrm{net}}={\stackrel{鈫�}{\mathrm{F}}}_{\mathrm{drag}}+{\stackrel{鈫�}{\mathrm{F}}}_{\mathrm{g}} \\ {\stackrel{鈫�}{\mathrm{F}}}_{\mathrm{net}}=0(\mathrm{for}\mathrm{terminal}\mathrm{speed}) \\ {\mathrm{v}}_{\text{terminal}}鈭�\mathrm{mg}=0 \\ {\mathrm{v}}_{\text{terminal}}=\frac{\mathrm{mg}}{\mathrm{b}}\left(\text{substitute}{\mathrm{F}}_{\text{net}}\right.\text{for terminal speed)} \end{gathered}$$

05

Part (c) Step 1: Given information

The radius is given by$\mathrm{r}=\sqrt{\frac{9{\mathrm{畏惫}}_{\mathrm{term}}}{2\mathrm{蚁驳}}}$

06

Part (c) Step 2: Calculations

We know from the problem's text that ${\mathrm{F}}_{\mathrm{drag}}=鈭�6\mathrm{蟺畏谤惫},$as a result

$$\begin{gathered} {\mathrm{F}}_{\text{drag}}鈥呪赌呪赌呪赌�=鈭�6\mathrm{蟺畏谤惫}=鈭�\mathrm{bv} \\ \mathrm{b}鈥呪赌呪赌呪赌�=6\mathrm{蟺畏谤}鈥呪赌呪赌呪赌�(\mathrm{from}\mathrm{previous}\mathrm{equation})鈥� \\ \mathrm{蚁}鈥呪赌呪赌呪赌�=\frac{\mathrm{m}}{\mathrm{V}}鈥呪赌呪赌呪赌�\text{(defintion of density)} \\ \mathrm{m}鈥呪赌呪赌呪赌�=\mathrm{蚁V} \\ \mathrm{V}鈥呪赌呪赌呪赌�=\frac{4}{3}{\mathrm{r}}^3\mathrm{蟺}鈥呪赌呪赌呪赌�\text{(express}\mathrm{m}) \\ \mathrm{m}鈥呪赌呪赌呪赌�=\mathrm{蚁}鈰�\frac{4}{3}{\mathrm{r}}^3\mathrm{蟺}鈥呪赌呪赌呪赌�\text{(volume of the droplet)}鈥呪赌� \end{gathered}$$

We may now use expressions for m and b in place of expressions for m and b.

$\begin{array}{r}{\mathrm{v}}_{\text{terminal}}=\frac{\mathrm{蚁}鈰�\frac{4}{3}{\mathrm{r}}^3\mathrm{蟺驳}}{6\mathrm{蟺畏谤}}\\ {\mathrm{v}}_{\text{terminal}}=\frac{2{\mathrm{r}}^2\mathrm{驳蚁}}{9\mathrm{畏}}\\ \mathrm{r}=\sqrt{\frac{9{\mathrm{畏惫}}_{\text{terminal}}}{2\mathrm{驳蚁}}}\left(\mathrm{express}\right)\end{array}$

07

Part (d) Step 1: Given information

The viscosity of air is$1.83脳{10}^{鈭�5}\mathrm{kg}/\mathrm{ms}$

08

Part (d) Step 2: Calculations

We may now utilise the expression from problem part a):

$$\begin{gathered} \mathrm{q}=\frac{\mathrm{mg}}{{\mathrm{E}}_0} \\ {\mathrm{E}}_0=\frac{\mathrm{螖痴}}{\mathrm{d}}\left(\mathrm{expressionforhomogenuselectricfieldbetweentwoparallelplates}\right) \\ \mathrm{m}={\mathrm{蚁}}_{\text{oil}}\mathrm{V} \\ \mathrm{m}={\mathrm{蚁}}_{\text{oil}}\frac{4}{3}{\mathrm{r}}^3\mathrm{蟺} \\ \mathrm{m}=\frac{4}{3}{\mathrm{蚁}}_{\text{oil}}\mathrm{蟺}{\left(\frac{9{\mathrm{畏惫}}_{\text{terminal}}}{2{\mathrm{驳蚁}}_{\text{oil}}}\right)}^{\frac{3}{2}} \\ \mathrm{q}=\frac{4}{3}{\mathrm{蚁}}_{\text{oil}}\mathrm{蟺}{\left(\frac{9{\mathrm{畏惫}}_{\text{terminal}}}{2{\mathrm{驳蚁}}_{\text{oil}}}\right)}^{\frac{3}{2}}鈰�\frac{\mathrm{gd}}{\mathrm{螖痴}}\text{(form c) part)}\left(\text{substitute}\mathrm{m}\text{and}{\mathrm{E}}_0\text{into expression for}\mathrm{q}\right.\text{)} \\ \mathrm{q}=\frac{4}{3}鈰�860鈰�\mathrm{蟺}{\left(\frac{9鈰�1.83鈰�{10}^{鈭�5}鈰�\frac{3鈰�{10}^{鈭�3}}{7.33}}{2鈰�9.8鈰�860}\right)}^{\frac{3}{2}}鈰�\frac{9.8鈰�0.01}{1177}\text{(substitute)} \\ \mathrm{q}=2.399鈰�{10}^{鈭�18}\mathrm{C} \end{gathered}$$

09

Part (e) Step 1: Given information

The gravitational and electric forces on the charged drops were equal as they dropped at a steady velocity.

10

Part (e) Step 2: Calculations

We can easily calculate the ratio between previous q and elementary charge e.

$$\begin{gathered} \mathrm{N}=\frac{\mathrm{q}}{\mathrm{e}} \\ \mathrm{N}=\frac{2.399鈰�{10}^{鈭�18}}{1.6鈰�{10}^{鈭�19}}\text{(substitute}\mathrm{q}\text{and}\mathrm{e}\text{)} \\ \mathrm{N}=15 \end{gathered}$$

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