Chapter 16, Problem 21P

To determine

The electric force on the $-0.60\mu \text{C}$ charge due to the other two charges.

Answer to Problem 21P

The electric force on the $-0.60\mu \text{C}$ charge due to the other two charges is $1.6\text{ N at }24°\text{ above the positive x-axis}$ .

Explanation of Solution

Write the Coulomb’s law.

$F=\frac{k\left|q_1\right|\left|q_2\right|}{r^2}$ (I)

Here, $F$ is the electric force, $k$ is a constant $q_1,q_2$ are the charges and $r$ is the distance between the charges

The force due to the $0.80\mu \text{C}$ charge is upward and that due to the $1.0\mu \text{C}$ is to the right.

Write the equation for the magnitude of the net force on $-0.60\mu \text{C}$ .

$F=\sqrt{F_x^2+F_y^2}$ (II)

Here, $F$ is the magnitude of the net force on $-0.60\mu \text{C}$ , $F_x$ is the magnitude of force on $-0.60\mu \text{C}$ due to $1.0\mu \text{C}$ and $F_y$ is the magnitude of force on $-0.60\mu \text{C}$ due to $0.80\mu \text{C}$

Write the equation for the direction of the net force on $-0.60\mu \text{C}$ .

$\theta ={\tan }^{-1}\frac{F_y}{F_x}$ (III)

Here, $\theta$ is the angle the net force on $-0.60\mu \text{C}$ makes with the horizontal

Conclusion:

The value of $k$ is $8.988\times {10}^9\text{ N}\cdot {\text{m}}^2{\text{/C}}^2$ .

Refer to the figure and find the force on $-0.60\mu \text{C}$ due to $0.80\mu \text{C}$ using equation (I).

$\begin{array}{c}{F}_y=\frac{\left(8.988\times {10}^9\text{ N}\cdot {\text{m}}^2{\text{/C}}^2\right)\left(0.80\mu \text{C}\left(\frac{1\text{ C}}{{10}^6\mu \text{C }}\right)\right)\left(0.60\mu \text{C}\left(\frac{1\text{ C}}{{10}^6\mu \text{C }}\right)\right)}{{\left(8.0\text{ cm}\left(\frac{1\text{ m}}{100\text{ cm}}\right)\right)}^2}\\ =\frac{\left(8.988\times {10}^9\text{ N}\cdot {\text{m}}^2{\text{/C}}^2\right)\left(0.80\times {10}^{-6}\text{ C}\right)\left(0.60\times {10}^{-6}\text{ C}\right)}{{\left(0.080\text{ m}\right)}^2}\\ =0.67\text{ N}\end{array}$

Refer to the figure and find the force on $-0.60\mu \text{C}$ due to $1.0\mu \text{C}$ using equation (I).

$\begin{array}{c}{F}_x=\frac{\left(8.988\times {10}^9\text{ N}\cdot {\text{m}}^2{\text{/C}}^2\right)\left(1.0\mu \text{C}\left(\frac{1\text{ C}}{{10}^6\mu \text{C }}\right)\right)\left(0.60\mu \text{C}\left(\frac{1\text{ C}}{{10}^6\mu \text{C }}\right)\right)}{\sqrt{{\left(10.0\text{ cm}\left(\frac{1\text{ m}}{100\text{ cm}}\right)\right)}^2-{\left(8.0\text{ cm}\left(\frac{1\text{ m}}{100\text{ cm}}\right)\right)}^2}}\\ =\frac{\left(8.988\times {10}^9\text{ N}\cdot {\text{m}}^2{\text{/C}}^2\right)\left(1.0\times {10}^{-6}\text{ C}\right)\left(0.60\times {10}^{-6}\text{ C}\right)}{\sqrt{{\left(0.100\text{ m}\right)}^2-{\left(0.080\text{ m}\right)}^2}}\\ =1.5\text{ N}\end{array}$

Substitute $0.67\text{ N}$ for $F_x$ and $1.5\text{ N}$ for $F_y$ in equation (II) to find $F$ .

$\begin{array}{c}F=\sqrt{{\left(0.67\text{ N}\right)}^2+{\left(1.5\text{ N}\right)}^2}\\ =1.6\text{ N}\end{array}$

Substitute $0.67\text{ N}$ for $F_x$ and $1.5\text{ N}$ for $F_y$ in equation (III) to find $\theta$ .

$\begin{array}{c}\theta ={\tan }^{-1}\frac{1.5\text{ N}}{0.67\text{ N}}\\ =24°\end{array}$

Therefore, the electric force on the $-0.60\mu \text{C}$ charge due to the other two charges is $1.6\text{ N at }24°\text{ above the positive x-axis}$ .