Chapter 16, Problem 22P

To determine

The electric force on the $+1.0\text{μC}$ charge due to the other two charges.

Answer to Problem 22P

The electric force on the $+1.0\text{μC}$ charge due to the other two charges is $\underset{\_}{1.2\text{N at }28°\text{ below the negative }x\text{-axis}}$.

Explanation of Solution

The arrangement of the charges is given in the figure. Let us represent $+1.0\text{μC}$ as $q_1$, $-0.60\text{μC}$ as $q_2$, and $+0.80\text{μC}$ as $q_3$.

Let us consider as the charges $q_1$ and $q_2$ be in the $x$-axis, with $q_1$ at the origin. Let $\widehat{i}$ and $\widehat{j}$ represents the unit vectors along $x$ and $y$ directions.

The use of Pythagoras theorem to the right triangle arrangement of the charges, the distance between the charges $q_1$ and $q_2$ is $6.0\text{cm}$. Hence the distance between $q_1$ and $q_2$ is $6.0\text{cm}$ ($0.06\text{m}$), and that between $q_1$ and $q_3$ is $10.0\text{cm}$ ($0.100\text{m}$).

According to superposition principle, the force on charge $q_1$ is the sum of forces due to $q_2$ and $q_3$.

Write the expression for the force on charge $q_1$ due to $q_2$.

${\overrightarrow{F}}_{12}=k{q}_1\left(\frac{q_2}{r_{12}^3}{\overrightarrow{r}}_{12}\right)$ (I)

Here, ${\overrightarrow{F}}_{\text{12}}$ is the force of charge $q_1$ due to $q_2$, $k$ is a constant ($k=8.988\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$), $r_{12}$ is the distance between $q_1$ and $q_2$, and ${\overrightarrow{r}}_{12}$ is the position vector of $q_2$ from the origin.

Write the expression for the force on charge $q_1$ due to $q_3$.

${\overrightarrow{F}}_{13}=k{q}_1\left(\frac{q_3}{r_{13}^3}{\overrightarrow{r}}_{13}\right)$ (II)

Here, ${\overrightarrow{F}}_{\text{13}}$ is the force of charge $q_1$ due to $q_3$, $r_{13}$ is the distance between $q_1$ and $q_3$, and ${\overrightarrow{r}}_{13}$ is the position vector of $q_3$ from the origin.

Write the expression for the total force on the charge $q_1$.

${\overrightarrow{F}}_{\text{total}}={\overrightarrow{F}}_{12}+{\overrightarrow{F}}_{13}$ (III)

Write the expression for the position vectors ${\overrightarrow{r}}_{12}$ and ${\overrightarrow{r}}_{13}$ from the figure.

${\overrightarrow{r}}_{12}=-0.060\widehat{i}+0.00\widehat{j}$ (IV)

${\overrightarrow{r}}_{13}=-0.060\widehat{i}+0.080\widehat{j}$ (V)

The magnitude of position vectors ${\overrightarrow{r}}_{12}$ and ${\overrightarrow{r}}_{13}$ is obtained from figure as, $r_{12}=0.060\text{m}$, and $r_{13}=0.100\text{m}$ respectively.

Use equation (IV) in (I).

${\overrightarrow{F}}_{12}=\frac{k{q}_1{q}_2}{r_{12}^3}\left(-0.060\widehat{i}+0.00\widehat{j}\right)$ (VI)

Use equation (V) in (II).

${\overrightarrow{F}}_{13}=\frac{k{q}_1{q}_3}{r_{13}^3}\left(-0.060\widehat{i}+0.080\widehat{j}\right)$ (VII)

Use equation (VI) and (VII) in (III).

$\begin{array}{c}{\overrightarrow{F}}_{\text{total}}=\frac{k{q}_1{q}_2}{r_{12}^3}\left(-0.060\widehat{i}+0.00\widehat{j}\right)+\frac{k{q}_1{q}_3}{r_{13}^3}\left(-0.060\widehat{i}+0.080\widehat{j}\right)\\ =k{q}_1\left[\frac{q_2}{r_{12}^3}\left(-0.060\widehat{i}+0.00\widehat{j}\right)+\frac{q_3}{r_{13}^3}\left(-0.060\widehat{i}+0.080\widehat{j}\right)\right]\\ =\left[-k{q}_1\left(\frac{q_2}{r_{12}^3}+\frac{q_3}{r_{13}^3}\right)\left(0.060\right)\right]\widehat{i}+\left[k{q}_1\frac{q_3}{r_{13}^3}\left(0.080\right)\right]\widehat{j}\end{array}$ (VIII)

Write the expression for the magnitude of the total force.

$F_{\text{total}}=\sqrt{F_x^2+F_y^2}$ (IX)

Here, $F_x$ and $F_y$ are the $x$ and $y$ components of the total force respectively.

Let $\theta$ be the angle that the resultant force makes below the negative $x$-axis. It is given by the expression,

$\theta ={\tan }^{-1}\left(\frac{F_y}{F_x}\right)$ (X)

Conclusion:

Substitute $8.988\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$, $+1.0\text{μC}$ for $q_1$, $-0.60\text{μC}$ for $q_2$, $+0.80\text{μC}$ for $q_3$, $0.060\text{m}$ for $r_{12}$, and $0.100\text{m}$ for $r_{13}$ in equation (VIII) to find ${\overrightarrow{F}}_{\text{total}}$.

$\begin{array}{c}{\overrightarrow{F}}_{\text{total}}=\left[-\left(8.988\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(+1.0\text{μC}\right)\left(\frac{-0.60\text{μC}}{{\left(0.060\text{m}\right)}^3}+\frac{+0.80\text{μC}}{{\left(0.100\text{m}\right)}^3}\right)\left(0.060\right)\right]\widehat{i}\\ +\left[\left(8.988\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(+1.0\text{μC}\right)\frac{+0.80\text{μC}}{{\left(0.100\text{m}\right)}^3}\left(0.080\right)\right]\widehat{j}\\ =\left[-\left(8.988\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(+1.0\text{μC}\times \frac{1\text{C}}{{10}^6\text{μC}}\right)\left(\frac{-0.60\text{μC}\times \frac{1\text{C}}{{10}^6\text{μC}}}{{\left(0.060\text{m}\right)}^3}+\frac{+0.80\text{μC}\times \frac{1\text{C}}{{10}^6\text{μC}}}{{\left(0.100\text{m}\right)}^3}\right)\left(0.060\right)\right]\widehat{i}\\ +\left[\left(8.988\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(+1.0\text{μC}\times \frac{1\text{C}}{{10}^6\text{μC}}\right)\frac{+0.80\text{μC}\times \frac{1\text{C}}{{10}^6\text{μC}}}{{\left(0.100\text{m}\right)}^3}\left(0.080\right)\right]\widehat{j}\\ =\left(1.1\text{N}\right)\widehat{i}+\left(0.58\text{N}\right)\widehat{j}\end{array}$

It is obtained that the magnitude of $x$ and $y$ components of the total force are $1.1\text{N}$ and $0.58\text{N}$ respectively.

Substitute $1.1\text{N}$ for $F_x$, and $0.58\text{N}$ for $F_y$ in equation (IX) to find $F_{\text{total}}$.

$\begin{array}{c}{F}_{\text{total}}=\sqrt{{\left(1.1\text{N}\right)}^2+{\left(0.58\text{N}\right)}^2}\\ =1.2\text{N}\end{array}$

Substitute $0.58\text{N}$ for $F_y$ and $1.1\text{N}$ for $F_x$ in equation (X) to find $\theta$.

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{0.58\text{N}}{1.1\text{N}}\right)\\ =28°\end{array}$

Therefore, the electric force on the $+1.0\text{μC}$ charge due to the other two charges is $\underset{\_}{1.2\text{N at }28°\text{ below the negative }x\text{-axis}}$.