Chapter 16, Problem 89P

To determine

The electric force on the chloride ion in the lower right-hand corner.

Answer to Problem 89P

The electric force on the chloride ion in the lower right-hand corner is $\underset{\_}{4\times {10}^{-13}\text{N at 60}°\text{ above the negative x-axis}}$.

Explanation of Solution

Write the equation of magnitude of the net electric force.

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

Here, $F$ is the magnitude of net force, $F_x$ is the x-component of electric force, $F_y$ is the y-component of electric force.

Write the equation of direction of the net electric force.

$\alpha =\frac{F_y}{F_x}$ (II)

Here, $\alpha$ is the direction of net force.

Write the equation of x-component of the net electric force.

$F_x=-\frac{k{q}^2}{x_1^2}+\frac{k{q}^2}{x_2^2}\cos \theta -\frac{k{q}^2}{x_3^2}\cos \varphi$ (III)

Here, $k$ is the coulomb’s constant, $q$ is the charge, $x_1,x_2,x_3$ are the distances.

Write the equation of y-component of the net electric force.

$F_y=-\frac{k{q}^2}{x_2^2}\sin \theta +\frac{k{q}^2}{x_3^2}\sin \varphi$ (IV)

Conclusion:

Substitute, $8.99\times {10}^9{\text{N-m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$, $2\times {10}^{-21}\text{C}$ for $q$, $0.5\text{nm}$ for $x_1$, $0.8\text{nm}$ for $x_2$, $0.3\text{nm}$ for $x_3$, $45°$ for $\theta$, $75°$ for $\varphi$ in equation (III)

$\begin{array}{c}{F}_x=-\left[\frac{\left(8.998\times {10}^9{\text{N-m}}^{\text{2}}{\text{/C}}^{\text{2}}\right){\left(2\times {10}^{-21}\text{C}\right)}^2}{{\left[\left(0.5\text{nm}\right)\left(\frac{1\times {10}^{-9}\text{m}}{1\text{nm}}\right)\right]}^2}\right]+\\ \left[\frac{\left(8.998\times {10}^9{\text{N-m}}^{\text{2}}{\text{/C}}^{\text{2}}\right){\left(2\times {10}^{-21}\text{C}\right)}^2}{{\left[\left(0.8\text{nm}\right)\left(\frac{1\times {10}^{-9}\text{m}}{1\text{nm}}\right)\right]}^2}\cos \left(45°\right)\right]-\\ \left[\frac{\left(8.998\times {10}^9{\text{N-m}}^{\text{2}}{\text{/C}}^{\text{2}}\right){\left(2\times {10}^{-21}\text{C}\right)}^2}{{\left[\left(0.3\text{nm}\right)\left(\frac{1\times {10}^{-9}\text{m}}{1\text{nm}}\right)\right]}^2}\cos \left(75°\right)\right]\\ =-2.07\times {10}^{-13}\text{N}\end{array}$

Substitute, $8.99\times {10}^9{\text{N-m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$, $2\times {10}^{-21}\text{C}$ for $q$, $0.8\text{nm}$ for $x_2$, $0.3\text{nm}$ for $x_3$, $45°$ for $\theta$, $75°$ for $\varphi$ in equation (IV)

$\begin{array}{c}{F}_y=-\left[\frac{\left(8.998\times {10}^9{\text{N-m}}^{\text{2}}{\text{/C}}^{\text{2}}\right){\left(2\times {10}^{-21}\text{C}\right)}^2}{{\left[\left(0.8\text{nm}\right)\left(\frac{1\times {10}^{-9}\text{m}}{1\text{nm}}\right)\right]}^2}\sin \left(45°\right)\right]-\\ \left[\frac{\left(8.998\times {10}^9{\text{N-m}}^{\text{2}}{\text{/C}}^{\text{2}}\right){\left(2\times {10}^{-21}\text{C}\right)}^2}{{\left[\left(0.3\text{nm}\right)\left(\frac{1\times {10}^{-9}\text{m}}{1\text{nm}}\right)\right]}^2}\sin \left(75°\right)\right]\\ =3.46\times {10}^{-13}\text{N}\end{array}$

Substitute, $-2.07\times {10}^{-13}\text{N}$ for $F_x$, $3.46\times {10}^{-13}\text{N}$ for $F_y$ in equation (I)

$\begin{array}{c}F=\sqrt{{\left(-2.07\times {10}^{-13}\text{N}\right)}^2+{\left(3.46\times {10}^{-13}\text{N}\right)}^2}\\ =4\times {10}^{-13}\text{N}\end{array}$

Substitute, $-2.07\times {10}^{-13}\text{N}$ for $F_x$, $3.46\times {10}^{-13}\text{N}$ for $F_y$ in equation (II)

$\begin{array}{c}\alpha ={\tan }^{-1}\left(\frac{\left(3.46\times {10}^{-13}\text{N}\right)}{\left(-2.07\times {10}^{-13}\text{N}\right)}\right)\\ =60°\text{ above the negative x-axis}\end{array}$

Thus, the electric force on the chloride ion in the lower right-hand corner is $\underset{\_}{4\times {10}^{-13}\text{N at 60}°\text{ above the negative x-axis}}$.