91Ó°ÊÓ

II 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 $E_0$will cause the droplet to be suspended motionless. Write an expression for the droplet's charge in terms of the suspending field $E_0$and the droplet's weight $mg$.

b. The field $E_0$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 $F_{\mathrm{drag}}=-bv$where $b$is a constant and $v$the droplet's 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 $v_{\text{term}}$in 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 rv where $\mathrm{η}$is the viscosity of the air. Use this and your answer to part b to show that a spherical droplet of density $\mathrm{ÒÏ}$falling with a terminal velocity $v_{\text{term}}$has a radius

$r=\sqrt{\frac{9\mathrm{η}{v}_{\text{term}}}{2\mathrm{ÒÏ}g}}$

d. Oil has a density $860\mathrm{kg}/{\mathrm{m}}^3$. An oil droplet is suspended between two plates $1.0\mathrm{cm}$apart by adjusting the potential difference between them to $1177\mathrm{V}$. When the voltage is removed, the droplet falls and quickly reaches constant speed. It is timed with a stopwatch, and falls $3.00\mathrm{mm}$in $7.33\mathrm{s}$. The viscosity of air is $1.83×{10}^{-5}\mathrm{kg}/\mathrm{ms}$. What is the droplet's charge?

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

Step by step solution

01

Part (a) Step 1: solution

a)We can start solution with Newton 's second law on a positive charge

$m\stackrel{→}{a}={\stackrel{→}{F}}_E+{\stackrel{→}{F}}_g$

$ma=q{E}_0-mg\text{expressions for electric and gravitational force)}$

$a=0\text{(charge suspended motionless)}$

$q{E}_0=mg\text{(substitue}a=0)$

$q=\frac{mg}{E_0}$(express } q

02

Part (b) Step 1:soluction

Now we can write Newton's second law with drag force:

$\begin{array}{r}{\stackrel{→}{F}}_{net}={\stackrel{→}{F}}_{drag}+{\stackrel{→}{F}}_g\\ {\stackrel{→}{F}}_{net}=0\text{(for terminal speed)}\\ b{v}_{\text{terminal}}-mg=0\text{(substitute}{F}_{net}\text{for terminal speed)}\\ v_{\text{terminal}}=\frac{mg}{b}\text{(express}{v}_{\text{terminal}}\text{)}\end{array}$

$\left(express{v}_{\text{terminal}}\right)$

03

Part (c) Step 1:soluction

From text of the problem we know that $F_{drag}=-6\mathrm{π}\mathrm{η}rv$, so we have:

$\begin{array}{r}{F}_{drag}=-6\mathrm{Ï€}\mathrm{η}rv=-bv\\ b=6\mathrm{Ï€}\mathrm{η}r\\ \mathrm{ÒÏ}=\frac{m}{V}\text{(from previous equation)}\\ m=\mathrm{ÒÏ}V\text{(defintion of density)}\\ V=\frac{4}{3}{r}^3\mathrm{Ï€}\text{(express}m\text{)}\\ m=\mathrm{ÒÏ}·\frac{4}{3}{r}^3\mathrm{Ï€}\text{(volume of the droplet)}\end{array}$

Now we can substitute expressions for $m$and $b$to expression for $v_{\text{terminal}}$

$$\begin{gathered} v_{\text{terminal}}=\frac{\mathrm{ÒÏ}·\frac{4}{3}{r}^3\mathrm{Ï€}g}{6\mathrm{Ï€}\mathrm{η}r} \\ {v}_{\text{terminal}}=\frac{2r^{2}g\mathrm{ÒÏ}}{9\mathrm{η}} \\ r=\sqrt{\frac{9\mathrm{η}{v}_{\text{terminal}}}{2g\mathrm{ÒÏ}}} \end{gathered}$$

$(expressr)$

04

Part (d) Step 1:soluction

Now we can use expression from part a) of the problem:

$$\begin{gathered} q=\frac{mg}{E_0} \\ {E}_0=\frac{\mathrm{Δ}V}{d}(\text{expression for homogenus electric field between two parallel plates)} \\ m={\mathrm{ÒÏ}}_{\text{oil}}V \\ m={\mathrm{ÒÏ}}_{\text{oil}}\frac{4}{3}{r}^3\mathrm{Ï€} \\ m=\frac{4}{3}{\mathrm{ÒÏ}}_{\text{oil}}\mathrm{Ï€}{\left(\frac{9\mathrm{η}{v}_{\text{terminal}}}{2g{\mathrm{ÒÏ}}_{\text{oil}}}\right)}^{\frac{3}{2}}(\text{form c) part)} \end{gathered}$$

$$\begin{gathered} q=\frac{4}{3}{\mathrm{ÒÏ}}_{\text{oil}}\mathrm{Ï€}{\left(\frac{9\mathrm{η}{v}_{\text{terminal}}}{2g{\mathrm{ÒÏ}}_{\text{oil}}}\right)}^{\frac{3}{2}}\left(\text{sutbotitute}m\text{and}{E}_0\text{into expression for}q\right) \\ q=\frac{4}{3}·860·\mathrm{Ï€}{\left(\frac{9·1.83·{10}^{-5}·\frac{3·{10}^{-3}}{7.33}}{2·9.8·860}\right)}^{\frac{3}{2}}·(\text{subbitititute)}1177\text{from c) part)} \\ q=2.399·{10}^{-18}\mathrm{C} \end{gathered}$$

05

Part (e) Step 1:soluction

Simply we can find ratio between previous $q$and elementary charge $e$:

$$\begin{gathered} N=\frac{q}{e} \\ N=\frac{2.399·{10}^{-18}}{1.6·{10}^{-19}} \\ N=15 \end{gathered}$$

role="math" localid="1650781163767" $(substituteqande)$

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