Hard Millikan Integral

Question

I tried to solve the Millikan integer, but I did not success. The initial integer is: $$m\frac{dv}{dt}= \frac{4}{3}\pi a^3 \rho g - qE - 6\pi \eta av$$ and I must prove show I to optain this: $$v= \frac{\frac{4}{3}\pi a^3 \rho g - qE} {6\pi \eta a}\left(1-e^\left(\frac{-6\pi \eta at}{m}\right)\right) $$

*** Just to make sure, all the "greek variables" are not variables in this equation; only $v$ and $t$.

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Those are my steps:

$$\int\frac{dv}{4/3 * \pi a^3 \rho g - qE - 6\pi \eta a v} = \int\frac{dt}{m}$$

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My substitution: $$u=4/3 * \pi a^3 \rho g - qE - 6\pi \eta a v$$ $$du= -6\pi \eta a (dv)$$ $$dv= \frac{-du}{6\pi \eta a}$$

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$$-\int\frac{du}{6\pi \eta a (u)} = \frac{t}{m} +c$$ $$\frac{-1}{6\pi \eta a} \ln \left\lvert 4/3 * \pi a^3 \rho g - qE - 6\pi \eta a v\right\rvert = \frac{t}{m} +c$$

$$ c=\frac{-1}{6\pi \eta a} \ln \left\lvert 4/3 * \pi a^3 \rho g - qE\right\rvert$$ We know that $t$ and $v$ are equal to 0.

$$-m \ln \left\lvert 4/3 * \pi a^3 \rho g - qE - 6\pi \eta a v\right\rvert= 6\pi \eta avt*\ln \left\lvert 4/3 * \pi a^3 \rho g - qE\right\rvert$$

$$ \ln \left\lvert \left( 4/3 * \pi a^3 \rho g - qE - 6\pi \eta a v \right)^{-m}\right\rvert= 6\pi \eta avt*\ln \left\lvert 4/3 * \pi a^3 \rho g - qE\right\rvert$$

$$\ln \left\lvert \frac{4/3 * \pi a^3 \rho g - qE}{\left( 4/3 * \pi a^3 \rho g - qE - 6\pi \eta a v \right)^{m}}\right\rvert = 6\pi \eta at$$

This is my last step, I don't know how to solve for v. Just in case, could you just make sure that I did not any mistakes so far and help me to solve for v. This should be the final answer: $$v= \frac{\frac{4}{3}\pi a^3 \rho g - qE} {6\pi \eta a}\left(1-e^\left(\frac{-6\pi \eta at}{m}\right)\right) $$

Answer 1

Writing the equation as $$m\frac{dv}{dt}=A-Bv$$ one can solve by separation of variables. Re-write as:

$$\frac{m}{A-Bv}\frac{dv}{dt}=1$$

Integrating $dt$ and making the substitution $u=A-Bv$:

$$\frac{-m}{B}\int\frac{du}{u}=t+\text{constant}$$

Continuing:

$$\ln(A-Bv)=-\frac{Bt}{m}+\text{constant}$$

$$v=\frac{1}{B}\left(A-ce^{-Bt/m}\right)$$

Since you have mentioned that $v(0)=0$, then this implies that $c=A$ and thus:

$$v=\frac{A}{B}\left(1-e^{-Bt/m}\right)$$

Finally note that $A=\frac{4}{3}\pi a^3\rho g-qE$, and $B=6\pi\eta a$.

Answer 2

$\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} m\,\totald{\mrm{v}\pars{t}}{t} & = {4 \over 3}\,\pi a^{3}\rho g - qE - 6\pi\eta a\mrm{v}\pars{t}\implies \pars{\totald{}{t} + {1 \over \tau}}\mrm{v}\pars{t} = {\tilde{v} \over \tau} \\[5mm] \mbox{where}\quad & \left\{\begin{array}{rcl} \ds{\tau} & \ds{\equiv} & \ds{m \over 6\pi\eta a} \\[2mm] \ds{\tilde{v}} & \ds{\equiv} & \ds{4\pi a^{3}\rho g/3 - qE \over 6\pi\eta a} \end{array}\right. \\[5mm] \implies & = \totald{\bracks{\expo{t/\tau}\mrm{v}\pars{t}}}{t} = {\tilde{v} \over \tau}\,\expo{t/\tau} \implies\expo{t/\tau}\mrm{v}\pars{t} - \expo{t_{0}/\tau}v\pars{t_{0}} = {\tilde{v} \over \tau}\,\int_{t_{0}}^{t}\expo{t'/\tau}\,\dd t' \end{align}

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\begin{align} \color{#f00}{\mrm{v}\pars{t}} & = v\pars{t_{0}}\expo{-\pars{t - t_{0}}/\tau} + {\tilde{v} \over \tau}\,\int_{t_{0}}^{t}\expo{-\pars{t - t'}/\tau}\,\dd t' = \color{#f00}{v\pars{t_{0}}\expo{-\pars{t - t_{0}}/\tau} + \tilde{v}\bracks{1 - \expo{-\pars{t - t_{0}}/\tau}}} \end{align}