Chapter 23, Problem 69AP

(a)

To determine

The electric charge at $(2.00\text{m},0)$

Answer to Problem 69AP

The electric field at $(2.00\text{m},0)$ is $24.2\widehat{\text{i}}\text{N}/\text{C}$.

Explanation of Solution

Conclusion:

Write the expression for the electric field.

    $\overrightarrow{E}=\left(E_1+E_2+E_3\right)\widehat{\text{i}}$                                                                                                   (I)

Here, s is the total electric field, $E_1$ is the field about the charge $-4.00\text{nC}$, $E_2$ is the field about the charge $5.00\text{nC}$, $E_3$ is the field about the charge $3.00\text{nC}$.

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 the position considered.

Substitute $\frac{kq}{r^2}$ in equation (I) and rewrite the equation.

    $\overrightarrow{E}=\left(\frac{k{q}_1}{r_1^2}+\frac{k{q}_2}{r_2^2}+\frac{k{q}_3}{r_3^2}\right)\widehat{\text{i}}$                                                                                         (III)

The following diagram shows the distance of the charges from $\left(2.00\text{m,}\text{0}\right)$.

Figure-(1)

Substitute $9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $k$, $-4.00\text{nC}$ for $q_1$, $2.500\text{m}$ $r_1$, $5.00\text{nC}$ for $q_2$, $2.00\text{m}$ for $r_2$, $3.00\text{nC}$ for $q_3$, $1.200\text{m}$ for $r_3$ in the equation (III) to get $E$.

    $\begin{array}{c}\overrightarrow{E}=\left(9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\left[\frac{-4.00\text{nC}}{{\left(2.500\text{m}\right)}^2}+\frac{5.00\text{nC}}{{\left(2.00\text{m}\right)}^2}+\frac{3.00\text{nC}}{{\left(1.200\text{m}\right)}^2}\right]\right)\widehat{\text{i}}\\ =\left(9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\left[-0.64\text{nC}/{\text{m}}^2+1.25\text{nC}/{\text{m}}^2+2.083\text{nC}/{\text{m}}^2\right]\right)\widehat{\text{i}}\\ \text{=}\left(9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\times 2.693\times {10}^{-9}\text{C}/{\text{m}}^2\right)\widehat{\text{i}}\\ =24.2\widehat{\text{i}}\text{N}/\text{C}\end{array}$

Therefore, the electric field at $\left(2.00\text{m,0}\right)$ is $24.2\widehat{\text{i}}\text{N}/\text{C}$.

(b)

To determine

The electric field at $\left(0,2.00\text{m}\right)$.

Answer to Problem 69AP

The resultant electric field is $9.4\text{ N/C}$ and the direction is $63.4°$ above $-x\text{axis}$.

Explanation of Solution

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

    ${\overrightarrow{E}}_{\text{x}}=\left[-\frac{k{q}_1}{r_1^2}\cos {\theta }_1+\frac{k{q}_2}{r_2^2}\cos {\theta }_2-\frac{k{q}_3}{r_3^2}\cos {\theta }_3\right]\widehat{\text{i}}$                                                         (IV)

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

    ${\overrightarrow{E}}_{\text{y}}=\left[\frac{k{q}_1}{r_1^2}\sin {\theta }_1+\frac{k{q}_2}{r_2^2}\sin {\theta }_2+\frac{k{q}_3}{r_3^2}\sin {\theta }_3\right]\overrightarrow{j}$                                                              (V)

Write the expression to calculate the electric field at $\left(0,2.00\text{m}\right)$.

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

Write the expression for the resultant electric field.

    $E_R=\sqrt{{\left({\overrightarrow{E}}_{\text{x}}\right)}^2+{\left({\overrightarrow{E}}_{\text{y}}\right)}^2}$                                                                                           (VII)

Here, $E_R$ is the resultant electric field.

Write the expression to calculate the direction.

  $\theta ={\tan }^{-1}\left(\frac{{\overrightarrow{E}}_{\text{y}}}{{\overrightarrow{E}}_{\text{x}}}\right)$                                                                                                    (VIII)

Here, $\theta$ is the angle.

Conclusion:

The following diagram shows the distance of the charges from $\left(0,2.00\text{m}\right)$.

Figure-(2)

From the figure-(2) calculate the distance of the charge from the position $\left(0,2.00\text{m}\right)$.

Consider the triangle $\angle ABO$,

    $\begin{array}{c}\tan {\theta }_{\text{1}}=\frac{2.00\text{m}}{0.500\text{m}}\\ {\theta }_{\text{1}}={\tan }^{-1}\left(4\right)\\ {\theta }_{\text{1}}=75.96°\end{array}$

Consider the triangle $\angle ABO$,

    $\begin{array}{c}\sin {\theta }_{\text{1}}=\frac{2.00\text{m}}{r_1}\\ r_1=\frac{2.00\text{m}}{\sin 75.964°}\\ =2.06\text{m}\end{array}$

Consider the triangle $\angle CBO$,

    $\begin{array}{c}\tan {\theta }_{\text{3}}=\frac{2.00\text{m}}{0.800\text{m}}\\ {\theta }_{\text{3}}={\tan }^{-1}\left(2.5\right)\\ {\theta }_{\text{3}}=68.19°\end{array}$

Consider the triangle $\angle CBO$,

    $\begin{array}{c}\sin {\theta }_{\text{3}}=\frac{2.00\text{m}}{r_3}\\ r_3=\frac{2.00\text{m}}{\sin 68.19°}\\ =2.15\text{m}\end{array}$

Substitute $9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $k$, $4.00\text{nC}$ for $q_1$, $2.06\text{m}$ $r_1$, $75.96°$ for ${\theta }_1$, $5.00\text{nC}$ for $q_2$, $2.00\text{m}$ for $r_2$, $90°$ for ${\theta }_2$, $3.00\text{nC}$ for $q_3$, $2.15\text{m}$ for $r_3$, $68.19°$ for ${\theta }_3$ in the equation (III) to get $E$.

    $\begin{array}{c}{\overrightarrow{E}}_{\text{x}}=\left[9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\left[\begin{array}{l}\frac{-4.00\text{nC}}{{\left(2.06\text{m}\right)}^2}\left(\cos 75.96°\right)+\\ \frac{5.00\text{nC}}{{\left(2.00\text{m}\right)}^2}\left(\cos 90°\right)-\frac{3.00\text{nC}}{{\left(2.15\text{m}\right)}^2}\left(\cos 68.19°\right)\end{array}\right]\right]\widehat{\text{i}}\\ =\left(9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\left[-0.228\text{nC}/{\text{m}}^2+0\text{nC}/{\text{m}}^2+-0.241\text{nC}/{\text{m}}^2\right]\right)\widehat{\text{i}}\\ \text{=}\left(9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\times \left(-0.469\text{nC}/{\text{m}}^2\times \frac{{10}^{-9}\text{C}}{\text{nC}}\right)\right)\widehat{\text{i}}\\ =-4.21\widehat{\text{i}}\text{N}/\text{C}\end{array}$

Substitute $9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $k$, $4.00\text{nC}$ for $q_1$, $2.06\text{m}$ $r_1$, $75.96°$ for ${\theta }_1$, $5.00\text{nC}$ for $q_2$, $2.00\text{m}$ for $r_2$, $90°$ for ${\theta }_2$, $3.00\text{nC}$ for $q_3$, $2.15\text{m}$ for $r_3$, $68.19°$ for ${\theta }_3$ in the equation (III) to get $E$.

    $\begin{array}{c}{\overrightarrow{E}}_{\text{y}}=\left[9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\left[\begin{array}{l}\frac{-4.00\text{nC}}{{\left(2.06\text{m}\right)}^2}\left(\sin 75.96°\right)+\frac{5.00\text{nC}}{{\left(2.00\text{m}\right)}^2}\left(\sin 90°\right)\\ +\frac{3.00\text{nC}}{{\left(2.15\text{m}\right)}^2}\left(\sin 68.19°\right)\end{array}\right]\right]\widehat{\text{j}}\\ =\left(9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\left[-0.914\text{nC}/{\text{m}}^2+1.25\text{nC}/{\text{m}}^2+0.60\text{nC}/{\text{m}}^2\right]\right)\widehat{\text{j}}\\ \text{=}\left(9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\times \left(0.842\text{nC}/{\text{m}}^2\times \times \frac{{10}^{-9}\text{C}}{\text{nC}}\right)\right)\widehat{\text{j}}\\ =8.42\widehat{\text{j}}\text{N}/\text{C}\end{array}$

Substitute $-4.21\widehat{\text{i}}\text{N}/\text{C}$ for ${\overrightarrow{E}}_{\text{x}}$ and $8.42\widehat{\text{j}}\text{N}/\text{C}$ for ${\overrightarrow{E}}_{\text{y}}$ in equation (VI) to solve for $\overrightarrow{E}$.

    $\begin{array}{c}\overrightarrow{E}=-4.21\widehat{\text{i}}\text{N}/\text{C}+8.42\widehat{\text{j}}\text{N}/\text{C}\\ =\left(-4.21\widehat{\text{i}}+8.42\widehat{\text{j}}\right)\text{N}/\text{C}\end{array}$

Then the resultant force is,

Substitute for $-4.21\widehat{\text{i}}\text{N}/\text{C}$ for ${\overrightarrow{E}}_{\text{x}}$ and ${\overrightarrow{E}}_{\text{y}}$ $8.42\widehat{\text{j}}\text{N}/\text{C}$ in  (VII).

    $\begin{array}{c}{E}_R=\sqrt{{\left(-4.21\widehat{\text{i}}\text{N}/\text{C}\right)}^2+{\left(8.42\widehat{\text{j}}\text{N}/\text{C}\right)}^2}\\ =9.4\text{ N/C}\end{array}$

Substitute for $-4.21\widehat{\text{i}}\text{N}/\text{C}$ for ${\overrightarrow{E}}_{\text{x}}$ and ${\overrightarrow{E}}_{\text{y}}$ $8.42\widehat{\text{j}}\text{N}/\text{C}$ in  (VIII).

Direction of the resultant force is,

    $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{8.42\widehat{\text{j}}\text{N}/\text{C}}{-4.21\widehat{\text{i}}\text{N}/\text{C}}\right)\\ =-63.4°\end{array}$

Therefore, resultant electric field is $9.4\text{ N/C}$ and the direction is $63.4°$ above $-x\text{axis}$.