Chapter 19, Problem 19P

Three point charges are arranged as shown in Figure P19.19. (a) Find the vector electric Field that the 6.00-nC and –3.00-nC charges together create at the origin. (b) Find the vector force on the 5.00-nC charge.

To determine

The electric field at the origin due to charges $6.00\text{nC}$ and $-3.00\text{nC}$.

Answer to Problem 19P

The electric field at the origin due to charges $6.00\text{nC}$ and $-3.00\text{nC}$ is $\underset{\_}{\overrightarrow{\text{E}}=-\left(5.99\times {10}^2\text{N}/\text{C}\right)\widehat{\text{i}}-\left(2.70\times {10}^3\text{N}/\text{C}\right)\widehat{\text{j}}}$.

Explanation of Solution

Figure 1 represents the electric field at the origin due to charges $6.00\text{nC}$ and $-3.00\text{nC}$.

Write the expression for the electric field at the origin due to $-3.00\text{nC}$ charge.

    ${\overrightarrow{\text{E}}}_1=\frac{k_e\left|q_1\right|}{r_1^2}\left(-\widehat{\text{j}}\right)$        (I)

Here, ${\overrightarrow{\text{E}}}_1$ is the electric field at the origin due to $-3.00\text{nC}$ charge, $k_e$ is the coulomb’s constant, $q_1$ is the charge, and $r_1$ is the distance from the origin to $-3.00\text{nC}$ charge.

Write the expression for the electric field at the origin due to $6.00\text{nC}$ charge.

    ${\overrightarrow{\text{E}}}_2=\frac{k_e\left|q_2\right|}{r_2^2}\left(-\widehat{\text{i}}\right)$        (II)

Here, ${\overrightarrow{\text{E}}}_1$ is the electric field at the origin due to $6.00\text{nC}$ charge, $k_e$ is the coulomb’s constant, $q_1$ is the charge, and $r_1$ is the distance from the origin to $6.00\text{nC}$ charge.

Write the expression for net electric field at the origin.

    $\overrightarrow{\text{E}}={\overrightarrow{\text{E}}}_1+{\overrightarrow{\text{E}}}_2$        (III)

Conclusion:

Substitute $8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $k_{\text{e}}$, $-3.00\text{nC}$ for $\left|q_1\right|$ $0.100\text{m}$ for $r_1$ in equation (I), to find ${\overrightarrow{\text{E}}}_1$.

    $\begin{array}{c}{\overrightarrow{\text{E}}}_1=\frac{\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\left|-3.00\text{nC}\times \frac{1\text{C}}{{10}^9\text{nC}}\right|}{\left(0.100\text{m}\right)}\left(-\widehat{\text{j}}\right)\\ =-\left(2.70\times {10}^3\text{N}/\text{C}\right)\widehat{\text{j}}\end{array}$        (IV)

Substitute $8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $k_{\text{e}}$, $6.00\text{nC}$ for $\left|q_2\right|$ $0.300\text{m}$ for $r_2$ in equation (II), to find ${\overrightarrow{\text{E}}}_2$.

    $\begin{array}{c}{\overrightarrow{\text{E}}}_2=\frac{\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\left|6.00\text{nC}\times \frac{1\text{C}}{{10}^9\text{nC}}\right|}{\left(0.300\text{m}\right)}\left(-\widehat{\text{i}}\right)\\ =-\left(5.99\times {10}^2\text{N}/\text{C}\right)\widehat{\text{i}}\end{array}$        (V)

Use equation (IV) and (V) in equation (III).

    $\overrightarrow{\text{E}}=-\left(2.70\times {10}^3\text{N}/\text{C}\right)\widehat{\text{j}}-\left(5.99\times {10}^2\text{N}/\text{C}\right)\widehat{\text{i}}$

Therefore, the electric field at the origin due to charges $6.00\text{nC}$ and $-3.00\text{nC}$ is $\underset{\_}{\overrightarrow{\text{E}}=-\left(5.99\times {10}^2\text{N}/\text{C}\right)\widehat{\text{i}}-\left(2.70\times {10}^3\text{N}/\text{C}\right)\widehat{\text{j}}}$.

To determine

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

Answer to Problem 19P

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

Explanation of Solution

From subpart (a) the electric field at the origin due to charges $6.00\text{nC}$ and $-3.00\text{nC}$ is $\overrightarrow{\text{E}}=-\left(5.99\times {10}^2\text{N}/\text{C}\right)\widehat{\text{i}}-\left(2.70\times {10}^3\text{N}/\text{C}\right)\widehat{\text{j}}$.

Write the expression for vector force on the charge.

    $\overrightarrow{\text{F}}=q\overrightarrow{\text{E}}$        (VI)

Here, $\overrightarrow{\text{F}}$ is the vector force on the charge, $q$ is the charge, and $\overrightarrow{\text{E}}$ is the electric field.

Conclusion:

Substitute $5.00\text{nC}$ for $q$, and $-\left(5.99\times {10}^2\text{N}/\text{C}\right)\widehat{\text{i}}-\left(2.70\times {10}^3\text{N}/\text{C}\right)\widehat{\text{j}}$ for $\overrightarrow{\text{E}}$ in equation (VI), to find $\overrightarrow{\text{F}}$.

    $\begin{array}{c}\overrightarrow{\text{F}}=\left(5.00\text{nC}\right)\left(-\left(5.99\times {10}^2\text{N}/\text{C}\right)\widehat{\text{i}}-\left(2.70\times {10}^3\text{N}/\text{C}\right)\widehat{\text{j}}\right)\text{N}/\text{C}\\ =\left(-3.00\widehat{\text{i}}-13.5\widehat{\text{j}}\right)\mu \text{N}\end{array}$

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