Chapter 21, Problem 86P

(a)

To determine

The charge on the sphere.

Explanation of Solution

Given:

The radius of the microsphere is $5.50\times {10}^7\text{m}$ .

The Magnitude of electric field is $6.00\times {10}^4\text{N}/\text{m}$ .

The magnitude of the drag force is $6\pi \eta rv$ .

The viscosity of the air is $1.8\times {10}^{-5}\text{N}\text{.s}/{\text{m}}^{\text{2}}$ .

The density of the polystyrene is $1.05\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$ .

Formula used:

Write the expression for the downward and upward force.

  $F_E-mg-F_d=m{a}_y$ ....... (1)

Here, $F_E$ is the electric force, $m$ is the mass, $g$ is acceleration due to gravity, $F_d$ is the drag force.

Write the expression for the force.

  $F=qE$

Here, $F$ is the force, $q$ is the charge and $E$ is the electric field.

Write the expression for the mass.

  $m=\rho V$

Here, $\rho$ is the density and $V$ is the volume.

Substitute $qE$ for $F_E$ , $\rho V$ for $m$ , $0$ for $a_y$ and $6\pi \eta rv$ for $F_d$ in equation (1).

  $qE-\rho Vg-6\pi \eta rv=0$ ....... (2)

Write the expression for the charge.

  $q=Ne$

Here, $q$ is the total charge, $N$ is the number of particles and $e$ is the charge on particle.

Write the expression for the volume of the sphere.

  $V=\frac{4}{3}\pi r^3$

Here, $r$ is the radius.

Substitute $Ne$ for $q$ and $\frac{4}{3}\pi r^3$ for $V$ in equation (2).

  $NeE-\rho \left(\frac{4}{3}\pi r^3\right)g-6\pi \eta rv=0$

Solve the above equation for $Ne$ .

  $Ne=\frac{\frac{4}{3}\pi r^3\rho g+6\pi \eta rv}{E}$ ....... (3)

Calculation:

Substitute $5.50\times {10}^7\text{m}$ for $r$ , $1.05\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ , $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ , $1.8\times {10}^{-5}\text{N}\text{.s}/{\text{m}}^{\text{2}}$ for $\eta$ , $1.16\times {10}^{-4}\text{m}/\text{s}$ for $v$ and $6.00\times {10}^4\text{N}/\text{m}$ for $E$ in equation (3).

  $\begin{array}{l}Ne=\frac{\frac{4}{3}\pi {\left(5.50\times {10}^7\text{m}\right)}^{3}1.05\times {10}^3\text{kg}/{\text{m}}^{\text{3}}\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)+6\pi 5.50\times {10}^7\left(1.8\times {10}^{-5}\text{N}\text{.s}/{\text{m}}^{\text{2}}\right)1.16\times {10}^{-4}\text{m}/\text{s}}{6.00\times {10}^4\text{N}/\text{m}}\\ Ne=4.8\times {10}^{-19}\text{C}\end{array}$

Conclusion:

Thus, the charge on the sphere is $4.8\times {10}^{-19}\text{C}$ .

(b)

To determine

The number of excess electron on the sphere.

Explanation of Solution

Given:

The radius of the microsphere is $5.50\times {10}^7\text{m}$ .

The Magnitude of electric field is $6.00\times {10}^4\text{N}/\text{m}$ .

The magnitude of the drag force is $6\pi \eta rv$ .

The viscosity of the air is $1.8\times {10}^{-5}\text{N}\text{.s}/{\text{m}}^{\text{2}}$ .

The density of the polystyrene is $1.05\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$ .

Formula used:

Write the expression for the downward and upward force.

  $F_E-mg-F_d=m{a}_y$

Here, $F_E$ is the electric force, $m$ is the mass, $g$ is acceleration due to gravity, $F_d$ is the drag force.

Write the expression for the force.

  $F=qE$

Here, $F$ is the force, $q$ is the charge and $E$ is the electric field.

Write the expression for the mass.

  $m=\rho V$

Here, $\rho$ is the density and $V$ is the volume.

Substitute $qE$ for $F_E$ , $\rho V$ for $m$ , $0$ for $a_y$ and $6\pi \eta rv$ for $F_d$ in equation (1).

  $qE-\rho Vg-6\pi \eta rv=0$

Write the expression for the charge.

  $q=Ne$

Here, $q$ is the total charge, $N$ is the number of particles and $e$ is the charge on particle.

Write the expression for the volume of the sphere.

  $V=\frac{4}{3}\pi r^3$

Here, $r$ is the radius.

Substitute $Ne$ for $q$ and $\frac{4}{3}\pi r^3$ for $V$ in equation (2).

  $NeE-\rho \left(\frac{4}{3}\pi r^3\right)g-6\pi \eta rv=0$

Solve the above equation for $N$ .

  $N=\frac{\frac{4}{3}\pi r^3\rho g+6\pi \eta rv}{Ee}$ ....... (4)

Calculation:

Substitute $5.50\times {10}^7\text{m}$ for $r$ , $1.05\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ , $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ , $1.8\times {10}^{-5}\text{N}\text{.s}/{\text{m}}^{\text{2}}$ for $\eta$ , $1.16\times {10}^{-4}\text{m}/\text{s}$ for $v$ and $6.00\times {10}^4\text{N}/\text{m}$ for $E$ and $1.602\times {10}^{-19}\text{C}$ for $e$ in equation (3).

  $\begin{array}{l}N=\frac{\frac{4}{3}\pi {\left(5.50\times {10}^7\text{m}\right)}^{3}1.05\times {10}^3\text{kg}/{\text{m}}^{\text{3}}\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)+6\pi 5.50\times {10}^7\left(1.8\times {10}^{-5}\text{N}\text{.s}/{\text{m}}^{\text{2}}\right)1.16\times {10}^{-4}\text{m}/\text{s}}{\left(1.602\times {10}^{-19}\text{C}\right)6.00\times {10}^4\text{N}/\text{m}}\\ N=3\end{array}$

Conclusion:

Thus, the number of excess electron on the sphere is 3.

(c)

To determine

The new terminal speed.

Explanation of Solution

Given:

The radius of the microsphere is $5.50\times {10}^7\text{m}$ .

The Magnitude of electric field is $6.00\times {10}^4\text{N}/\text{m}$ .

The magnitude of the drag force is $6\pi \eta rv$ .

The viscosity of the air is $1.8\times {10}^{-5}\text{N}\text{.s}/{\text{m}}^{\text{2}}$ .

The density of the polystyrene is $1.05\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$ .

Formula used:

Write the expression for the forces when electric field is upward.

  $F_d-F_E-mg=0$ ....... (5)

Here, $F_E$ is the electric force, $m$ is the mass, $g$ is acceleration due to gravity, $F_d$ is the drag force.

Write the expression for the force.

  $F=eE$

Here, $F$ is the force, $e$ is the charge and $E$ is the electric field.

Write the expression for the mass.

  $m=\rho V$

Here, $\rho$ is the density and $V$ is the volume.

Substitute $qE$ for $F_E$ , $\rho V$ for $m$ , $\frac{4}{3}\pi r^3$ for $V$ and $6\pi \eta rv$ for $F_d$ in equation (1).

  $6\pi \eta rv-NeE-\left(\frac{4}{3}\pi r^3\right)\rho g=0$

Solve the above equation for $v$ .

  $v=\frac{NeE+\left(\frac{4}{3}\pi r^3\right)\rho g}{6\pi \eta r}$ ....... (6)

Calculation:

Substitute $5.50\times {10}^7\text{m}$ for $r$ , $1.05\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ , $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ , $1.8\times {10}^{-5}\text{N}\text{.s}/{\text{m}}^{\text{2}}$ for $\eta$ , $3$ for $N$ , $6.00\times {10}^4\text{N}/\text{m}$ for $E$ and $1.602\times {10}^{-19}\text{C}$ for $e$ in equation (3).

  $\begin{array}{l}v=\frac{3\left(1.602\times {10}^{-19}\text{C}\right)6.00\times {10}^4\text{N}/\text{m}+\left(\frac{4}{3}\pi {\left(5.50\times {10}^7\text{m}\right)}^3\right)\left(1.05\times {10}^3\text{kg}/{\text{m}}^{\text{3}}\right)9.81\text{m}/{\text{s}}^{\text{2}}}{6\pi \left(1.8\times {10}^{-5}\text{N}\text{.s}/{\text{m}}^{\text{2}}\right)5.50\times {10}^7\text{m}}\\ v=0.19\text{mm}/\text{s}\end{array}$

Conclusion:

Thus, the new terminal speed is $0.19\text{mm}/\text{s}$ .