Chapter 16, Problem 114P

To determine

The terminal speed of charged rain drop.

Answer to Problem 114P

Terminal speed is $6.8\text{m}/\text{s}$.

Explanation of Solution

Radius of rain drop is $1.0\text{mm}$, charge is $2\text{nC}$, electric field is $2000\text{N}/\text{C}$, terminal speed of uncharged drop is $6.5\text{m}/\text{s}$, and the drag force is $F_d=b{v}^2$.

Write the relation between buoyant force, drag force, and force of gravity on an uncharged drop at equilibrium.

$F_B+F_d=F_g$ . (I)

Here, $F_B$ is the buoyant force, $F_d$ is the drag force, and $F_g$ is the weight of drop.

Write the equation for $F_B$.

$F_B={\rho }_{air}{V}_{drop}g$ . (II)

Here, ${\rho }_{air}$ is the density of air, $V_{drop}$ is the volume of drop, and $g$ is the gravitational acceleration.

Write the equation for $F_d$.

$F_d=b{v}^2$ . (III)

Here, $b$ is the drag constant and $v$ is the terminal speed of uncharged drop.

Write the equation for $F_g$.

$F_g={\rho }_{drop}{V}_{drop}g$ . (IV)

Here, ${\rho }_{drop}$ is the density of water.

Rewrite equation (I) by substituting the above relations for $F_B$,$F_d$ and $F_g$.

$\begin{array}{c}{\rho }_{air}{V}_{drop}g+b{v}^2={\rho }_{drop}{V}_{drop}g\\ b=\frac{\left({\rho }_{drop}-{\rho }_{air}\right)V_{drop}g}{v^2}\end{array}$

Write the relation between buoyant forces, drag force, force of gravity, force due to electric field, and on an uncharged drop at equilibrium.

$F_B+F_d=F_g+F_E$ . (V)

Here, $F_E$ is the force on drop due to electric field.

Write the equation for $F_d$ of charged drop.

$F_d=b{v}_t^2$ . (VI)

Here, $v_t$ is the terminal speed of charged drop.

Write the equation for $F_E$.

$F_E=qE$ . (VII)

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

Rewrite equation (V) by substituting equations (II), (IV), (VI), and (VII).

$\begin{array}{c}{\rho }_{air}{V}_{drop}g+b{v}_t^2={\rho }_{drop}{V}_{drop}g+qE\\ v_t=\sqrt{\frac{\left({\rho }_{drop}-{\rho }_{air}\right)V_{drop}g+qE}{b}}\end{array}$

Rewrite the above equation by substituting $\frac{\left({\rho }_{drop}-{\rho }_{air}\right)V_{drop}g}{v^2}$ for $b$.

$\begin{array}{c}{v}_t=\sqrt{\frac{\left({\rho }_{drop}-{\rho }_{air}\right)V_{drop}g+qE}{\left(\frac{\left({\rho }_{drop}-{\rho }_{air}\right)V_{drop}g}{v^2}\right)}}\\ =v\sqrt{1+\frac{qE}{\left({\rho }_{drop}-{\rho }_{air}\right)V_{drop}g}}\end{array}$

Write the equation for $V_{drop}$.

$V_{drop}=\frac{4}{3}\pi r^3$

Here, $r$ is the radius of drop.

Rewrite the equation for $v$ by substituting the above relation.

$v_t=v\sqrt{1+\frac{qE}{\left({\rho }_{drop}-{\rho }_{air}\right)\left(\frac{4}{3}\pi r^3\right)g}}$

Conclusion:

Substitute $6.5\text{m}/\text{s}$ for $v$, $2\text{nC}$ for $q$, $2000\text{N}/\text{C}$ for $E$, $1000\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_d$, $1.20\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_a$, $1.0\text{mm}$ for $r$, and $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ in the above equation to find $v_t$.

$\begin{array}{c}{v}_t=\left(6.5\text{m}/\text{s}\right)\sqrt{1+\frac{\left(2\text{nC}\right)\left(2000\text{N}/\text{C}\right)}{\left(1000\text{kg}/{\text{m}}^{\text{3}}-1.20\text{kg}/{\text{m}}^{\text{3}}\right)\left(\frac{4}{3}\pi {\left(1.0\text{mm}\right)}^3\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)}}\\ =\left(6.5\text{m}/\text{s}\right)\left(1.046\right)\\ =6.8\text{m}/\text{s}\end{array}$

Therefore, the terminal speed is $6.8\text{m}/\text{s}$.