Chapter 17, Problem 125P

An air ionizer fillers particles of dust, pollen, and other allergens from the air using electric forces. In one type of ionizer (see diagram), a stream of air is drawn in with a speed of 3.0 m/s. The air passes through a fine, highly charged wire mesh that transfers electric charge to the particles. Then the air passes through parallel “collector” plates that admet the charged particles and trap them in a filter. Consider a dust particle of radius 6.0 μm, mass 2.0 × 10^–13 kg, and charge l000e. The plates are 10 cm long and are separated by a distance of 10 cm. (a) Ignoring drag forces, what would be the minimum potential difference between the plates to ensure that the particle gets trapped by the filter? (b) At what speed would the particle be moving relative to the stream of air just before hitting the filter? (c) Calculate the viscous drag force on the particle when moving at the speed found in (b). (d) Is it realistic to ignore drag? Taking drag into consideration, is the minimum potential difference larger or smaller than the answer to (a)?

To determine

The minimum potential difference between the plates to ensure that the particle gets trapped by the filter.

Answer to Problem 125P

The minimum potential difference between the plates to ensure that the particle gets trapped by the filter is $220\text{ V}$ .

Explanation of Solution

Write the equation for the time taken by the dust particle to pass through the length of the collector plates.

  $\Delta t=\frac{\Delta x}{v_x}$        (I)

Here, $\Delta t$ is the time taken by the dust particle to pass through the length of the collector plates, $\Delta x$ is the length of the collector plates and $v_x$ is the horizontal speed of the particle.

Write the equation of motion in vertical direction.

  $\Delta y=v_i\Delta t+\frac{1}{2}{a}_y{\left(\Delta t\right)}^2$

Here, $\Delta y$ is the vertical displacement, $v_i$ is the initial speed of the particle in vertical direction and $a_y$ is the acceleration of the particle in the vertical direction.

The initial speed of the particle in the vertical direction is zero.

Substitute $0$ for $v_i$ in the above equation and rewrite for $a_y$ .

  $\begin{array}{c}\Delta y=0+\frac{1}{2}{a}_y{\left(\Delta t\right)}^2\\ a_y=\frac{2\Delta y}{{\left(\Delta t\right)}^2}\end{array}$        (II)

Put equation (I) in equation (II).

  $\begin{array}{c}{a}_y=\frac{2\Delta y}{{\left(\frac{\Delta x}{v_x}\right)}^2}\\ =\frac{2v_x^2\Delta y}{{\left(\Delta x\right)}^2}\end{array}$        (III)

Write the equation for the net force on the particle in the vertical direction.

  $F=m{a}_y$

Here, $F$ is the net force on the particle in the vertical direction and $m$ is the mass of the particle.

Rewrite the above equation for $a_y$ .

  $a_y=\frac{F}{m}$        (IV)

The net force in the vertical direction is equal to the electric force.

Write the relationship between the net force and the electric force.

  $F=F_e$

Here, $F_e$ is the electric force.

Put the above equation in equation (IV).

  $a_y=\frac{F_e}{m}$        (V)

Write the equation for the electric force.

  $F_e=qE$

Here, $q$ is the charge on the particle and $E$ is the electric field.

Write the equation for the electric field.

  $E=\frac{\Delta V}{\Delta y}$        (VI)

Here, $\Delta V$ is the potential difference between the plates.

Put the above equation in equation (VI).

  $F_e=\frac{q\Delta V}{\Delta y}$        (VII)

Put equation (VII) in equation (V).

  $a_y=\frac{q\Delta V}{m\Delta y}$        (VIII)

Equate equations (III) and (VIII) and rewrite it for $\Delta V$ .

  $\begin{array}{c}\frac{2v_x^2\Delta y}{{\left(\Delta x\right)}^2}=\frac{q\Delta V}{m\Delta y}\\ \Delta V=\frac{2m{v}_x^2{\left(\Delta y\right)}^2}{q{\left(\Delta x\right)}^2}\end{array}$

Substitute $1000e$ for $q$ in the above equation.

  $\Delta V=\frac{2m{v}_x^2{\left(\Delta y\right)}^2}{1000e{\left(\Delta x\right)}^2}$        (IX)

Here, $e$ is the magnitude of the electronic charge.

Conclusion:

The value of $e$ is $1.602\times {10}^{-19}\text{ C}$ .

Substitute $2.0\times {10}^{-13}\text{ kg}$ for $m$ , $3\text{ m/s}$ for $v_x$ , $1\text{ cm}$ for $\Delta y$ , $1.602\times {10}^{-19}\text{ C}$ for $e$ and $10\text{ cm}$ for $\Delta x$ in equation (IX) to find $\Delta V$ .

  $\begin{array}{c}\Delta V=\frac{2\left(2.0\times {10}^{-13}\text{ kg}\right){\left(3\text{ m/s}\right)}^2{\left(1\text{ cm}\frac{1\text{ m}}{100\text{ cm}}\right)}^2}{1000\left(1.602\times {10}^{-19}\text{ C}\right){\left(10\text{ cm}\frac{1\text{ m}}{100\text{ cm}}\right)}^2}\\ =220\text{ V}\end{array}$

Therefore, the minimum potential difference between the plates to ensure that the particle gets trapped by the filter is $220\text{ V}$ .

To determine

The speed at which the particle would be moving relative to the stream of air just before hitting the filter.

Answer to Problem 125P

The speed at which the particle would be moving relative to the stream of air just before hitting the filter is $0.60\text{ m/s}$ .

Explanation of Solution

Write the equation of motion in vertical direction.

  $v_f=v_i+a_y\Delta t$

Here, $v_f$ is the speed at which the particle would be moving relative to the stream of air just before hitting the filter.

Substitute $0$ for $v_i$ in the above equation.

  $\begin{array}{c}{v}_f=0+a_y\Delta t\\ =a_y\Delta t\end{array}$

Put equation (II) in the above equation.

  $\begin{array}{c}{v}_f=\frac{2\Delta y}{{\left(\Delta t\right)}^2}\Delta t\\ =\frac{2\Delta y}{\Delta t}\end{array}$

Put equation (I) in the above equation.

  $\begin{array}{c}{v}_f=\frac{2\Delta y}{\frac{\Delta x}{v_x}}\\ =\frac{2v_x\Delta y}{\Delta x}\end{array}$        (X)

Conclusion:

Substitute $3\text{ m/s}$ for $v_x$ , $1\text{ cm}$ for $\Delta y$ and $10\text{ cm}$ for $\Delta x$ in equation (X) to find $v_f$ .

  $\begin{array}{c}{v}_f=\frac{2\left(3\text{ m/s}\right)\left(1\text{ cm}\frac{1\text{ m}}{100\text{ cm}}\right)}{\left(10\text{ cm}\frac{1\text{ m}}{100\text{ cm}}\right)}\\ =0.60\text{ m/s}\end{array}$

Therefore, the speed at which the particle would be moving relative to the stream of air just before hitting the filter is $0.60\text{ m/s}$ .

To determine

The viscous drag force on the particle when moving at the speed found in part (b).

Answer to Problem 125P

The viscous drag force on the particle when moving at the speed found in part (b) is $1.2\text{ nN}$ .

Explanation of Solution

Write the equation for the drag force.

  $F_D=6\pi \eta rv$        (XI)

Here, $F_D$ is the drag force, $\eta$ is the coefficient of viscosity, $r$ is the radius of the particle and $v$ is the speed.

Conclusion:

The viscosity of air is $1.8\times {10}^{-5}\text{ Pa}\cdot \text{s}$ .

Substitute $1.8\times {10}^{-5}\text{ Pa}\cdot \text{s}$ for $\eta$ , $6.0\text{ μm}$ for $r$ and $0.60\text{ m/s}$ for $v$ in equation (IX) to find $F_D$ .

  $\begin{array}{c}{F}_D=6\pi \left(1.8\times {10}^{-5}\text{ Pa}\cdot \text{s}\right)\left(6.0\text{ μm}\frac{1\text{ m}}{{10}^6\text{ μm}}\right)\left(0.60\text{ m/s}\right)\\ =1.2\times {10}^{-9}\text{ N}\\ =1.2\times {10}^{-9}\text{ N}\frac{1\text{ nN}}{{10}^{-9}\text{ N}}\\ =1.2\text{ nN}\end{array}$

Therefore, the viscous drag force on the particle when moving at the speed found in part (b) is $1.2\text{ nN}$ .

To determine

Whether it is realistic to ignore the drag and whether the minimum potential difference larger or smaller than the answer to part (a) when drag is considered.

Answer to Problem 125P

It is not realistic to ignore the drag and the minimum potential difference should be larger when drag is considered.

Explanation of Solution

It is realistic to ignore the drag if the drag force is much smaller than the electric force.

Substitute $1000e$ for $q$ in equation (VII).

  $F_e=\frac{1000e\Delta V}{\Delta y}$        (XII)

Conclusion:

Substitute $1.602\times {10}^{-19}\text{ C}$ for $e$ , $220\text{ V}$ for $\Delta V$ and $1\text{ cm}$ for $\Delta y$ in equation (XII) to find $F_e$ .

  $\begin{array}{c}{F}_e=\frac{1000\left(1.602\times {10}^{-19}\text{ C}\right)\left(220\text{ V}\right)}{1\text{ cm}\frac{1\text{ m}}{100\text{ cm}}}\\ =3.5\times {10}^{-12}\text{ N}\\ =3.5\times {10}^{-12}\text{ N}\frac{1\text{ pN}}{{10}^{-12}\text{ N}}\\ =3.5\text{ pN}\end{array}$

The value of $F_D$ is $1.2\text{ nN}$ so that $F_e<<F_D$ . This implies it is not realistic to ignore the drag force. When drag force is considered, the minimum potential difference must be larger than that found in part (a).

Therefore, it is not realistic to ignore the drag and the minimum potential difference should be larger when drag is considered.