Chapter 23, Problem 35P

(a)

To determine

The vector electric field created by the $6.00\text{nC}$ and $-3.00\text{nC}$ charges together at the origin.

Answer to Problem 35P

The vector electric field that the $6.00\text{nC}$ and $-3.00\text{nC}$ charges together create at the origin is $\left[-0.599\widehat{\text{i}}-2.70\widehat{\text{j}}\right]\text{kN}/\text{C}$.

Explanation of Solution

The following figure represents the forces acting on the charges.

Figure-(1)

Here, $F_1$ is the electric force acting on the charge of $5.00\text{nC}$ by the charge of $6.00\text{nC}$ and $F_2$ is the electric force acting on the charge $5.00\text{nC}$ by the charge $-3.00\text{nC}$.

Write the expression for the electric field at origin by the charge $q_1$ in the negative $x$ axis.

    $\overrightarrow{E_1}=\frac{k{q}_1}{r_1^2}\left(-\widehat{\text{i}}\right)$

Here, $\overrightarrow{E_1}$ is the electric field at origin by the charge $q_1$, $r_1$ is the distance of the charge $q_1$ from the origin, $k$ is the electrostatic constant. The negative sign indicates that the electric field is in negative $x$ axis direction.

Write the constant value for the Coulomb constant $k$.

    $k=8.9876\times {\text{10}}^9\text{N}.{\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}$

Write the expression for the electric field at origin by the charge $q_2$ in the negative $y$ axis.

    $\overrightarrow{E_2}=\frac{k{q}_2}{r_2^2}\left(-\widehat{\text{j}}\right)$

Here, $\overrightarrow{E_2}$ is the electric field at origin by the charge $q_2$, $r_2$ is the distance of the charge $q_2$ from the origin, $k$ is the electrostatic constant. The negative sign indicates that the electric field is in negative $y$ axis direction.

Write the expression for the total vector electric field at origin by the charges $q_1$ and $q_2$.

    $\overrightarrow{E}=\overrightarrow{E_1}+\overrightarrow{E_2}$                                                                                                                (I)

Here, the $\overrightarrow{E_1}$ is the electric field at origin by the charge $q_1$ and $\overrightarrow{E_2}$ is the electric field at origin by the charge $q_2$.

Write the expression to convert the units for charge $q_1$ from $\text{nC}$ to $\text{C}$.

    $q_1=6.00\text{nC}$                                                                                                             (II)

Write the expression to convert the units for charge $q_2$ from $\text{nC}$ to $\text{C}$.

    $q_2=\left|-3.00\text{nC}\right|$                                                                                                        (III)

Substitute $\frac{k{q}_1}{r_1^2}\left(-\widehat{\text{i}}\right)$ for $\overrightarrow{E_1}$ and $\frac{k{q}_2}{r_2^2}\left(-\widehat{\text{j}}\right)$ for $\overrightarrow{E_2}$ in the equation (I) to calculate $\overrightarrow{E}$.

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

Conclusion:

Convert the units for the charge $q_1$ from $\text{nC}$ to $\text{C}$ in equation (II).

    $\begin{array}{c}{q}_1=6.00\text{nC}\left(\frac{1\text{C}}{1\times {10}^9\text{nC}}\right)\\ =6.00\times {10}^{-9}\text{C}\end{array}$

Convert the units for the charge $q_2$ from $\text{nC}$ to $\text{C}$ in equation (III).

    $\begin{array}{c}{q}_2=-3.00\text{nC}\left(\frac{1\text{C}}{1\times {10}^9\text{nC}}\right)\\ =-3.00\times {10}^{-9}\text{C}\end{array}$

Substitute $6.00\times {10}^{-9}\text{C}$ for $q_1$, $-3.00\times {10}^{-9}\text{C}$ for $q_2$, $0.300\text{m}$ for $r_1$ and $0.100\text{m}$ for $r_2$ and $8.9876\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $k$ in equation (IV) to solve $\overrightarrow{E}$.

    $\begin{array}{c}\overrightarrow{E}=\left[\begin{array}{c}\frac{\left(8.9876\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\left(6.00\times {10}^{-9}\text{C}\right)}{{\left(0.300\text{m}\right)}^2}\left(-\widehat{\text{i}}\right)+\\ \frac{\left(8.9876\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\left(3.00\times {10}^{-9}\text{C}\right)}{{\left(0.100\text{m}\right)}^2}\left(-\widehat{\text{j}}\right)\end{array}\right]\\ =\left[599.17\left(-\widehat{\text{i}}\right)+2696.28\left(-\widehat{\text{j}}\right)\right]\text{N}/\text{C}\\ =\left[0.599\times {10}^3\left(-\widehat{\text{i}}\right)+2.70\times {10}^3\left(-\widehat{\text{j}}\right)\right]\text{N}/\text{C}\end{array}$

Further simplify the above equation.

    $\begin{array}{c}\overrightarrow{E}=\left[-0.599\times {10}^3\widehat{\text{i}}-2.70\times {10}^3\widehat{\text{j}}\right]\text{N}/\text{C}\\ =\left[-0.599\times {10}^3\widehat{\text{i}}-2.70\times {10}^3\widehat{\text{j}}\right]\text{N}/\text{C}\left(\frac{{\text{10}}^{-3}\text{kN}/\text{C}}{\text{1}\text{N}/\text{C}}\right)\\ =\left[-0.599\widehat{\text{i}}-2.70\widehat{\text{j}}\right]\text{kN}/\text{C}\end{array}$

Therefore, the vector electric field that the $6.00\text{nC}$ and $-3.00\text{nC}$ charges together create at the origin is $\left[-0.599\widehat{\text{i}}-2.70\widehat{\text{j}}\right]\text{kN}/\text{C}$.

(b)

To determine

The vector force on the $5.00\text{nC}$ charge.

Answer to Problem 35P

The vector force on the $5.00\text{nC}$ charge is $\left(-3.00\widehat{\text{i}}-13.5\widehat{\text{j}}\right)\mu \text{N}$.

Explanation of Solution

Write the expression for the relation between the electric force and the electric field.

    $\overrightarrow{F}=q\overrightarrow{E}$                                                                                                                   (IV)

Here, $q$ is the charge on the origin, $\overrightarrow{E}$ is the electric field at the origin due to the charges $6.00\text{nC}$ and $-3.00\text{nC}$.

Write the expression to convert the units for charge $q_1$ from $\text{nC}$ to $\text{C}$.

    $q=5.00\text{nC}$                                                                                                              (V)

Conclusion:

Convert the units for the charge $q$ from $\text{nC}$ to $\text{C}$ in equation (V).

    $\begin{array}{c}q=5.00\text{nC}\left(\frac{1\text{C}}{1\times {10}^9\text{nC}}\right)\\ =5.00\times {10}^{-9}\text{C}\end{array}$

Substitute $5.00\times {10}^{-9}\text{C}$ for $q$ and $\left[-0.599\widehat{\text{i}}-2.70\widehat{\text{j}}\right]\times {10}^3\text{N}/\text{C}$ for $\overrightarrow{E}$ in equation (IV) to calculate $\overrightarrow{F}$.

    $\begin{array}{c}\overrightarrow{F}=5.00\times {10}^{-9}\text{C}\times \left[-0.599\widehat{\text{i}}-2.70\widehat{\text{j}}\right]\times {10}^3\text{N}/\text{C}\\ =\left[-\left(5.00\times {10}^{-9}\right)\times \left(0.599\widehat{\text{i}}\right)-\left(5.00\times {10}^{-9}\text{C}\right)\times \left(2.70\widehat{\text{j}}\right)\right]\times {10}^3\text{N}\\ =\left[\left(-2.99\widehat{\text{i}}-13.48\widehat{\text{j}}\right)\times {10}^{-6}\text{N}\right]\left(\frac{1\mu \text{N}}{1\times {10}^{-6}\text{N}}\right)\\ \approx \left(-3.00\widehat{\text{i}}-13.5\widehat{\text{j}}\right)\mu \text{N}\end{array}$

Therefore, the vector force on the $5.00\text{nC}$ charge is $\left(-3.00\widehat{\text{i}}-13.5\widehat{\text{j}}\right)\mu \text{N}$.