An electrostatic dust precipitator that is installed in a factory smokestack includes a straight metal wire of length L=0.8 mthat is charged approximately uniformly with a total charge $\mathbf{Q}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{}\mathbf{C}$ . A speck of coal dust (which is mostly carbon) is near the wire, far from both ends of the wire; the distance from the wire to the speck is d=1.5 cm . Carbon has an atomic mass of 12( 6protons and 6neutrons in the nucleus). A careful measurement of the polarizability of a carbon atom gives the value

$\mathbf{\propto }\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{96}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{40}}\frac{\mathbf{C}\mathbf{·}\mathbf{m}}{\mathbf{N}\mathbf{/}\mathbf{C}}$

(a) Calculate the initial acceleration of the speck of coal dust, neglecting gravity. Explain your steps clearly. Your answer must be expressed in terms of$\mathbf{Q}\mathbf{,}\mathbf{}\mathbf{L}\mathbf{,}\mathbf{}\mathbf{d}\mathbf{,}$and $\mathbf{\propto }$. You can use other quantities in your calculations, but your final result must not include them. Don’t put numbers into your calculation until the very end, but then show the numerical calculation that you carry out on your calculator. It is convenient to use the “binomial expansion” that you may have learned in calculus, that${\left(1+\epsilon \right)}^{\mathbf{n}}\mathbf{\approx }\mathbf{1}\mathbf{+}\mathbf{n\epsilon }$is $\mathbf{\epsilon }\mathbf{\ll }\mathbf{1}$. Note thatcan be negative. (b) If the speck of coal dust were initially twice as far from the charged wire, how much smaller would be the initial acceleration of the speck?

The given data is listed below as-

• The length of the straight metal wire is, $L=0.8\mathrm{m}$.
• The charge of the metal wire is, $Q=0.4\times {10}^{-7}\mathrm{C}$.
• The wire’s distance from the speck is, $d=1.5\mathrm{cm}\left(\frac{{10}^{-2}\mathrm{m}}{1\mathrm{cm}}\right)=0.015\mathrm{m}$.
• The atomic mass of carbon is 12 .
• The value of the polarizability of the carbon atom is, $\propto =1.96\times {10}^{-40}\frac{\mathrm{C}·\mathrm{m}}{\mathrm{N}/\mathrm{C}}$.

The electric field’s magnitude establishes proportional to the charge and shows an inversely proportional relationship with the specific distance square.

In this problem, the electric field equation gives the initial acceleration of the speck.

The equation of the magnitude of the electric field can be expressed as:

$E=k\frac{Q}{d\sqrt{d^2+{\left(L/2\right)}^2}}$

Here,$E$is the magnitude of the electric field, $k$is the electric field constant, $Q$is the charge induced, $L$is the length of the wire and $d$is radial distance.

The equation of the dipole moment is expressed as:

$$\begin{gathered} p=\propto E \\ =\propto k\frac{Q}{d\sqrt{d^2+{\left(L/2\right)}^2}} \end{gathered}$$

Here, $p$is the dipole moment, $\propto$is the value of the polarizability of the carbon atom and $E$is the magnitude of the electric field, $k$is the electric field constant, $Q$is the charge induced, $L$is the length of the wire, and $d$is radial distance.

The equation of the force of the rod on the dipole is expressed as:

$$\begin{gathered} F=k\frac{pQ}{d^2\sqrt{d^2+{\left(L/2\right)}^2}} \\ =k\frac{Q}{d^2\sqrt{d^2+{\left(L/2\right)}^2}}\times \propto k\frac{Q}{d\sqrt{d^2+{\left(L/2\right)}^2}} \end{gathered}$$

Here, $F$is the force of the rod on the dipole, $p$is the dipole moment, $Q$is the charge induced, $L$is the length of the wire andd is radial distance.

The equation of the mass of the carbon atom is expressed as:

$m_c=12m$

Here, $m_c$is the mass of the carbon atom and is the atomic weight of the carbon atom and atomic mass of the carbon is 12 .

Substitute $1.66\times {10}^{-27}\mathrm{kg}$for $m$in the above equation.

$$\begin{gathered} m_c=12\times 1.66\times {10}^{-27}\mathrm{kg} \\ =1.99\times {10}^{-26}\mathrm{kg} \end{gathered}$$

The equation of the initial acceleration is given as:

$a_1=\frac{F}{m_c}$

Here, $a_1$is the initial acceleration, $F$is the force of the rod on the dipole and $m_c$is the mass of the carbon atom.

Substitute all the values in the above equation.

$$\begin{gathered} a_1=\frac{1}{m_c}\times k\frac{Q}{d^2\sqrt{d^2+{\left(L/2\right)}^2}}\times \propto k\frac{Q}{d\sqrt{d^2+{\left(L/2\right)}^2}} \\ =\frac{1}{m_c}\times k^2\times \frac{\propto }{d^3}\frac{Q}{d^3+{\left(L/2\right)}^2} \end{gathered}$$ …(i)

Substitute all the values in the above equation.

$$\begin{gathered} {\mathrm{a}}_1\approx \frac{1}{1.99\times {10}^{-26}\mathrm{kg}}{\left(8.99\times {10}^9\mathrm{N}·{\mathrm{m}}^2/{\mathrm{C}}^2\right)}^2\frac{1.96\times {10}^{-40}{\displaystyle \frac{\mathrm{C}·\mathrm{m}}{\mathrm{N}/\mathrm{C}}}}{{\left(0.015\mathrm{m}\right)}^3}·\frac{{\left(0.4\times {10}^{-7}\mathrm{C}\right)}^2}{{\left(0.015\mathrm{m}\right)}^2+{\left(0.8\mathrm{m}/2\right)}^2} \\ \approx \frac{1}{1.99\times {10}^{-26}\mathrm{kg}}{\left(8.99\times {10}^9\mathrm{N}·{\mathrm{m}}^2/{\mathrm{C}}^2\right)}^2\frac{1.96\times {10}^{-40}{\displaystyle \frac{\mathrm{C}·\mathrm{m}}{\mathrm{N}/\mathrm{C}}}}{{\left(0.015\mathrm{m}\right)}^3}·\frac{{\left(0.4\times {10}^{-7}\mathrm{C}\right)}^2}{\left(0.016\mathrm{m}\right)} \\ \approx \frac{\left(8.08\times {10}^{19}\mathrm{N}·{\mathrm{m}}^2/{\mathrm{C}}^2\right)}{1.99\times {10}^{-26}\mathrm{kg}}\times \frac{1.96\times {10}^{-40}{\displaystyle \frac{\mathrm{C}·\mathrm{m}}{\mathrm{N}/\mathrm{C}}}}{{\left(0.015\mathrm{m}\right)}^3}·1\times {10}^{-14}{\mathrm{C}}^2/{\mathrm{m}}^2 \\ \approx 4.06\times {10}^{45}{\mathrm{N}}^2·{\mathrm{m}}^4/{\mathrm{C}}^4·\mathrm{kg}\times 5.807\times {10}^{-35}\frac{{\mathrm{C}}^2}{\mathrm{N}·{\mathrm{m}}^2}1\times {10}^{-14}{\mathrm{C}}^2/{\mathrm{m}}^2 \end{gathered}$$

Hence, further simplified as,

$$\begin{gathered} a_1\approx 4.06\times {10}^{45}{\mathrm{N}}^2·{\mathrm{m}}^4/{\mathrm{C}}^2·\mathrm{kg}\times 5.807\times {10}^{-35}\frac{{\mathrm{C}}^2}{\mathrm{N}·\mathrm{m}}·1\times {10}^{-14}{\mathrm{C}}^2/{\mathrm{m}}^2 \\ \approx 2.35\times {10}^{11}\frac{{\mathrm{N}}^2·{\mathrm{m}}^4}{{\mathrm{C}}^2·\mathrm{kg}}\times \frac{{\mathrm{C}}^2}{\mathrm{N}·\mathrm{m}}·1\times {10}^{-14}{\mathrm{C}}^2/{\mathrm{m}}^2 \\ \approx 2.35\times {10}^{11}\mathrm{N}·{\mathrm{m}}^4/{\mathrm{C}}^2·\mathrm{kg}·1\times {10}^{-14}{\mathrm{C}}^2/{\mathrm{m}}^2 \\ \approx 2.35\times {10}^{-3}\mathrm{N}/\mathrm{kg} \end{gathered}$$

Hence, further simplified as,

$$\begin{gathered} a_1\approx 2.35\times {10}^{-3}\mathrm{N}/\mathrm{kg} \\ \approx 2.35\times {10}^{-3}\frac{\mathrm{N}}{\mathrm{kg}}\left(\frac{1\mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2}{1\mathrm{N}}\right) \\ \approx 2.35\times {10}^{-3}\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Thus, the initial acceleration of the speck of coal dust is $2.35\times {10}^{-3}\mathrm{m}/{\mathrm{s}}^2$.

The equation (i) represents the initial acceleration of the speck of coal dust.

If the wire’s distance from the speck is made twice, then the distance will be:

$$\begin{gathered} d=0.015\mathrm{m}\times 2 \\ =0.03\mathrm{m} \end{gathered}$$

The equation of the initial acceleration is given as:

$a_2=\frac{F}{m_c}$

Here, $a_2$is the initial acceleration, $F$is the force of the rod on the dipole and $m_c$is the mass of the carbon atom.

Substitute all the values in the above equation.

localid="1656990943461" $$\begin{gathered} a_2=\frac{1}{m_c}\times k\frac{Q}{d^2\sqrt{d^2+{\left(L/2\right)}^2}}\times \propto k\frac{Q}{d\sqrt{d^2+{\left(L/2\right)}^2}} \\ =\frac{1}{m_c}\times k^2\times \frac{\propto }{d^3}\frac{Q^2}{d^2+{\left(L/2\right)}^2} \end{gathered}$$ …(ii)

Substitute the values in the equation (ii).

$$\begin{gathered} a_2\approx \frac{1}{1.99\times {10}^{-26}\mathrm{kg}}\left(8.99\times {10}^9\mathrm{N}·{\mathrm{m}}^2/{\mathrm{C}}^2\right)\frac{1.99\times {10}^{-40}{\displaystyle \frac{\mathrm{C}.\mathrm{m}}{\mathrm{N}/\mathrm{C}}}}{{\left(0.03\mathrm{m}\right)}^3}·\frac{{\left(0.4\times {10}^{-7}\mathrm{C}\right)}^2}{{\left(0.03\mathrm{m}\right)}^2+{\left(0.8\mathrm{m}/2\right)}^2} \\ \approx \frac{1}{1.99\times {10}^{-26}\mathrm{kg}}\left(8.99\times {10}^9\mathrm{N}·{\mathrm{m}}^2/{\mathrm{C}}^2\right)\frac{1.99\times {10}^{-40}{\displaystyle \frac{\mathrm{C}.\mathrm{m}}{\mathrm{N}/\mathrm{C}}}}{{\left(0.03\mathrm{m}\right)}^3}·\frac{{\left(0.4\times {10}^{-7}\mathrm{C}\right)}^2}{{\left(0.16\mathrm{m}\right)}^2} \\ \approx \frac{8.08\times {10}^{19}{\mathrm{N}}^2·{\mathrm{m}}^4/{\mathrm{C}}^4}{1.99\times {10}^{-26}\mathrm{kg}}\times \frac{1.99\times {10}^{-40}{\displaystyle \frac{\mathrm{C}.\mathrm{m}}{\mathrm{N}/\mathrm{C}}}}{{\left(0.03\mathrm{m}\right)}^3}·1\times {10}^{-14}{\mathrm{C}}^2/{\mathrm{m}}^2 \\ \approx 4.06\times {10}^{45}{\mathrm{N}}^2·{\mathrm{m}}^4/{\mathrm{C}}^4·\mathrm{kg}\times 7.25\times {10}^{-35}\frac{{\mathrm{C}}^2}{\mathrm{N}·{\mathrm{m}}^2}·1\times {10}^{-14}{\mathrm{C}}^2/{\mathrm{m}}^2 \end{gathered}$$

Hence, further simplified as,

$$\begin{gathered} a_2\approx 4.06\times {10}^{45}{\mathrm{N}}^2·{\mathrm{m}}^4/{\mathrm{C}}^4·\mathrm{kg}\times 7.25\times {10}^{-35}\frac{{\mathrm{C}}^2}{\mathrm{N}·{\mathrm{m}}^2}·1\times {10}^{-14}{\mathrm{C}}^2/{\mathrm{m}}^2 \\ \approx 2.93\times {10}^{10}\frac{{\mathrm{N}}^2·{\mathrm{m}}^4}{{\mathrm{C}}^4·\mathrm{kg}}\times \frac{{\mathrm{C}}^2}{\mathrm{N}·{\mathrm{m}}^2}·1\times {10}^{-14}{\mathrm{C}}^2/{\mathrm{m}}^2 \\ \approx 2.93\times {10}^{10}\mathrm{N}·{\mathrm{m}}^4/{\mathrm{C}}^4·\mathrm{kg}·1\times {10}^{-14}{\mathrm{C}}^2/{\mathrm{m}}^2 \\ \approx 2.93\times {10}^{10}\mathrm{N}/\mathrm{kg} \end{gathered}$$

Hence, further simplified as,

$$\begin{gathered} a_2\approx 2.93\times {10}^{10}\mathrm{N}/\mathrm{kg} \\ \approx 2.93\times {10}^{-4}\frac{\mathrm{N}}{\mathrm{kg}}\left(\frac{1\mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2}{1\mathrm{N}}\right) \\ \approx 2.93\times {10}^{-4}\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Hence, the equation to find the smaller initial acceleration is:

$a=a_1-a_2$

Substituting $2.93\times {10}^{-4}\mathrm{m}/\mathrm{s}$for $a_2$and $2.35\times {10}^{-3}\mathrm{m}/{\mathrm{s}}^2$for $a_1$in the above equation.

$$\begin{gathered} a=2.35\times {10}^{-3}\mathrm{m}/{\mathrm{s}}^2-2.93\times {10}^{-4}\mathrm{m}/{\mathrm{s}}^2 \\ =2.1\times {10}^{-3}\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Thus, the initial acceleration of the speck will be smaller by $2.1\times {10}^{-3}\mathrm{m}/{\mathrm{s}}^2$.