Chapter 23, Problem 72AP

(a)

To determine

The magnitude of the electric force exerted on the charge at $q_1$ $\left(0,0\right)$.

Answer to Problem 72AP

The magnitude of the electric force exerted on the charge at $q_1$ is $40.85\text{N}$.

Explanation of Solution

Write the expression for the total electric field.

    $\overrightarrow{E}={\overrightarrow{E}}_{\text{x}}+{\overrightarrow{E}}_{\text{y}}$                                                                                                                (I)

Here, $\overrightarrow{E}$ is the total electric field, ${\overrightarrow{E}}_{\text{x}}$ is the electric field resolved in $x$ axis, ${\overrightarrow{E}}_{\text{y}}$ is the electric field resolved in $y$ axis.

Write the expression for the electric field.

    $E=\frac{kq}{r^2}$                                                                                                                      (II)

Here, $k$ is the constant, $q$ is the electric charge, $r$ is the distance from $q_1$.

Rewrite the equation (II) for resolving the forces along $x$ direction.

    ${\overrightarrow{E}}_{\text{x}}=\left[\frac{k{q}_2}{r_2^2}\cos {\theta }_2+\frac{k{q}_3}{r_3^2}\cos {\theta }_3+\frac{k{q}_4}{r_4^2}\cos {\theta }_4\right]\widehat{i}$                                                           (III)

Here, $q_2$ is the charge at B, ${\theta }_2$ is the angle made by the $q_2$ with $q_1$, $r_2$ is the distance of $q_2$ form $q_1$, $q_3$ is the charge at C, ${\theta }_3$ is the angle made by the $q_3$ with $q_1$, $r_3$ is the distance of $q_3$ form $q_1$, $q_4$ is the charge at D, ${\theta }_4$ is the angle made by the $q_4$ with $q_1$, $r_4$ is the distance of $q_4$ form $q_1$.

Rewrite the equation (II) for resolving the forces along $y$ direction.

    ${\overrightarrow{E}}_{\text{y}}=\left[\frac{k{q}_2}{r_2^2}\sin {\theta }_2+\frac{k{q}_3}{r_3^2}\sin {\theta }_3+\frac{k{q}_4}{r_4^2}\sin {\theta }_4\right]\widehat{j}$                                                            (IV)

Write the expression for the electric force.

    $F=qE$                                                                                                                     (V)

Conclusion:

The following diagram shows the distance of the charges from $\left(0,0\right)$.

Figure-(1)

From the figure-(1) calculate the distance of $q_3$ from the position $\left(0,0\right)$.

Consider the triangle $\Delta CAD$.

    $\begin{array}{c}\tan {\theta }_{\text{3}}=\frac{15.0\text{cm}}{60.0\text{cm}}\\ {\theta }_{\text{3}}={\tan }^{-1}\left(0.25\right)\\ {\theta }_{\text{3}}=14.036°\end{array}$

Consider the triangle $\Delta CAD$.

    $\begin{array}{c}\sin {\theta }_{\text{3}}=\frac{15.0\text{cm}}{r_3}\\ r_3=\frac{15.0\text{cm}}{\sin \left(14.036°\right)}\\ r_3=61.847\text{cm}\end{array}$

Substitute $9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $k$, $+10.0\mu \text{C}$ for $q_2$, $15.0\text{cm}$ for $r_2$, $90°$ for ${\theta }_2$, $+10.0\mu \text{C}$ for $q_3$, $61.847\text{cm}$ for $r_3$, $14.036°$ for ${\theta }_3$, $+10.0\mu \text{C}$ for $q_4$, $60.0\text{cm}$ $r_4$, $0°$ for ${\theta }_4$,in the equation (III) to find ${\overrightarrow{E}}_{\text{x}}$.

    $\begin{array}{c}{\overrightarrow{E}}_{\text{x}}=\left[9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\left(+10.0\times {10}^{-6}\text{C}\right)\left(\begin{array}{l}\frac{1}{{\left(15.0\text{cm}\right)}^2}\cos 90°+\\ \frac{1}{{\left(61.847\text{cm}\right)}^2}\cos 14.036°+\\ \frac{1}{{\left(60.0\text{cm}\right)}^2}\cos 0°\end{array}\right)\right]\widehat{i}\\ =\left[90000\text{N}\cdot {\text{m}}^2\left(\begin{array}{l}\frac{1}{{\left(0.15\text{m}\right)}^2}\cos 90°+\\ \frac{1}{{\left(0.618\text{cm}\right)}^2}\cos 14.036°+\frac{1}{{\left(0.60\text{cm}\right)}^2}\cos 0°\end{array}\right)\right]\widehat{i}\\ =\left[90000\text{N}\cdot {\text{m}}^2\left(\begin{array}{l}01/{\text{m}}^2+\\ 2.5401/{\text{m}}^2+2.7781/{\text{m}}^2\end{array}\right)\right]\widehat{i}\end{array}$

Calculate further.

    $\begin{array}{c}{\overrightarrow{E}}_{\text{x}}\text{=}\left(9\times {10}^4\text{N}\cdot {\text{m}}^2\times 5.318\text{1}/{\text{m}}^2\right)\widehat{i}\\ =478.62\times {10}^3\widehat{i}\text{N}\end{array}$

Substitute $9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $k$, $+10.0\mu \text{C}$ for $q_2$, $15.0\text{cm}$ for $r_2$, $90°$ for ${\theta }_2$, $+10.0\mu \text{C}$ for $q_3$, $61.847\text{cm}$ for $r_3$, $14.036°$ for ${\theta }_3$, $+10.0\mu \text{C}$ for $q_4$, $60.0\text{cm}$ $r_4$, $0°$ for ${\theta }_4$ in equation (III) to find ${\overrightarrow{E}}_{\text{y}}$.

    $\begin{array}{c}{\overrightarrow{E}}_{\text{y}}=\left[9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\left(+10.0\times {10}^{-6}\text{C}\right)\left[\begin{array}{l}\frac{1}{{\left(15.0\text{cm}\right)}^2}\sin 90°+\\ \frac{1}{{\left(61.847\text{cm}\right)}^2}\sin 14.036°+\\ \frac{1}{{\left(60.0\text{cm}\right)}^2}\sin 0°\end{array}\right]\right]\widehat{j}\\ =\left[90000\text{N}\cdot {\text{m}}^2\left[\begin{array}{l}\frac{1}{{\left(0.15\text{m}\right)}^2}\sin 90°+\\ \frac{1}{{\left(0.618\text{cm}\right)}^2}\sin 14.036°+\frac{1}{{\left(0.60\text{cm}\right)}^2}\sin 0°\end{array}\right]\right]\widehat{j}\\ =\left(90000\text{N}\cdot {\text{m}}^2\left[\begin{array}{l}44.4441/{\text{m}}^2+\\ 0.6351/{\text{m}}^2+01/{\text{m}}^2\end{array}\right]\right)\widehat{j}\end{array}$

Calculate further.

    $\begin{array}{c}{\overrightarrow{E}}_{\text{y}}\text{=}\left(9\times {10}^4\text{N}\cdot {\text{m}}^2\times 45.079\text{1}/{\text{m}}^2\right)\widehat{j}\\ =4057.11\times {10}^3\widehat{j}\text{N}\end{array}$

Substitute $478.62\times {10}^3\widehat{i}\text{N}$ for ${\overrightarrow{E}}_{\text{x}}$, $4057.11\times {10}^3\widehat{j}\text{N}$ for ${\overrightarrow{E}}_{\text{y}}$ in the equation (I) to find $\overrightarrow{E}$.

    $\begin{array}{c}\overrightarrow{E}=478.62\times {10}^3\widehat{i}\text{N}+4057.11\times {10}^3\widehat{j}\text{N}\\ =\left(478.62\widehat{i}+4057.11\widehat{j}\right)\times {10}^3\text{N}\end{array}$

Substitute $\left(478.62\widehat{i}+4057.11\widehat{j}\right)\times {10}^3\text{N}$ for $\overrightarrow{E}$ and $+10.0\times {10}^{-6}\text{C}$ for $q$ in the equation (V) to find $\overrightarrow{F}$.

    $\begin{array}{c}\overrightarrow{F}=\left[+10.0\times {10}^{-6}\text{C}\right]\left[\left(478.62\widehat{i}+4057.11\widehat{j}\right)\times {10}^3\text{N}\right]\\ =4.786\widehat{i}+40.57\widehat{j}\text{NC}\end{array}$

Calculate the magnitude of the electric force.

    $\begin{array}{c}F=\sqrt{\left({\left(4.786\widehat{i}\right)}^2+{\left(40.57\widehat{j}\right)}^2\right)}\text{N}\\ \text{=}\sqrt{22.906+1645.93}\text{N}\\ =4\text{0}\text{.85}\text{N}\end{array}$

Therefore, the magnitude of the electric force exerted on the charge is $40.85\text{N}$.

(b)

To determine

The direction of the electric force exerted on the charge at $q_1$ $\left(0,0\right)$.

Answer to Problem 72AP

The direction of the electric force exerted on the charge at $q_1$ is $6.728°$.

Explanation of Solution

Write the expression for the direction of the electric force.

    $\tan \alpha =\frac{A}{B}$                                                                                                               (VI)

Here, $\alpha$ is the direction of the field, $A$ is the coefficient of the electric force in $x$ axis, $B$ is the coefficient of the electric force in $y$ axis.

Conclusion:

Substitute $4.786$ for $A$, $40.57$ for $B$ in the equation (VI) to find $\alpha$.

    $\begin{array}{c}\tan \alpha =\frac{4.786}{40.57}\\ \alpha ={\tan }^{-1}\left(\frac{4.786}{40.57}\right)\\ \alpha ={\tan }^{-1}\left(0.118\right)\\ \alpha =6.728°\end{array}$

Therefore, the direction of the electric force exerted on the charge is $6.728°$.