Chapter 21, Problem 43P

(a)

To determine

The magnitude and direction of electric field at $\left(-1.0\text{m,}0\right)$ .

Explanation of Solution

Given:

The magnitude of the first point charge is $-5.0\text{μC}$ .

The position of the first point charge is $\left(4.0\text{m},-\text{2}\text{m}\right)$ .

The magnitude of the second point charge is $1\text{2}\text{μC}$ .

The position of the second point charge is $\left(-1.0\text{m},0\text{m}\right)$ .

Formula used:

Write the expression for the resultant electric field at point $P$ .

  ${\overrightarrow{E}}_P={\overrightarrow{E}}_1+{\overrightarrow{E}}_2$ ..................(1)

Here, ${\overrightarrow{E}}_P$ is the electric field at point $P$ , ${\overrightarrow{E}}_1$ is the electric field due to the $q_1$ charge and ${\overrightarrow{E}}_2$ is the electric field due to $q_2$ charge.

Write the expression for the electric field due to charge $q_1$ .

  ${\overrightarrow{E}}_1=\frac{k{q}_1}{r_{1,P}^2}{\widehat{r}}_{1,P}$ ..................(2)

Here, $k$ is constant, $q_1$ is the first charge and $r_{1,P}$ is the distance between the point $P$ and the first charge.

Write the expression for the electric field due to charge $q_2$ .

  ${\overrightarrow{E}}_2=\frac{k{q}_2}{r_{2,P}^2}{\widehat{r}}_{2,P}$ ..................(3)

Here, $k$ is constant, $q_2$ is the first charge and $r_{2,P}$ is the distance between the point $P$ and the second charge.

Calculation:

Substitute $8.988\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $k$ , $-5.0\text{μC}$ for $q_1$ , $5.0\text{m}$ and $2\text{m}$ for $r_{1,P}$ in equation (1).

  $\begin{array}{l}{\overrightarrow{E}}_1=\frac{8.988\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}\left(-5.0\text{μC}\right)}{{\left(5.0\text{m}\right)}^2+{\left(2.0\text{m}\right)}^2}\left[\frac{\left(-5.0\text{m}\right)\widehat{i}+\left(2.0\text{m}\right)\widehat{j}}{\sqrt{{\left(5.0\text{m}\right)}^2+{\left(2.0\text{m}\right)}^2}}\right]\\ {\overrightarrow{E}}_1=-1.55\times {10}^3\text{N}/\text{C}\left(-0.928\widehat{i}+0.371\widehat{j}\right)\\ {\overrightarrow{E}}_1=\left(1.44\text{kN}/\text{C}\right)\widehat{i}-\left(0.575\text{kN}/\text{C}\right)\widehat{j}\end{array}$

Substitute $8.988\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $k$ , $12.0\text{μC}$ for $q_1$ , $2.0\text{m}$ and $2.0\text{m}$ for $r_{1,P}$ in equation (2).

  $\begin{array}{l}{\overrightarrow{E}}_2=\frac{8.988\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}\left(12.0\text{μC}\right)}{{\left(2.0\text{m}\right)}^2+{\left(2.0\text{m}\right)}^2}\left[\frac{\left(-2.0\text{m}\right)\widehat{i}+\left(-2.0\text{m}\right)\widehat{j}}{\sqrt{{\left(5.0\text{m}\right)}^2+{\left(2.0\text{m}\right)}^2}}\right]\\ {\overrightarrow{E}}_2=13.5\times {10}^3\text{N}/\text{C}\left(0.707\widehat{i}-0.707\widehat{j}\right)\\ {\overrightarrow{E}}_2=\left(-9.54\text{kN}/\text{C}\right)\widehat{i}-\left(9.54\text{kN}/\text{C}\right)\widehat{j}\end{array}$

Substitute $\left(1.44\text{kN}/\text{C}\right)\widehat{i}-\left(0.575\text{kN}/\text{C}\right)\widehat{j}$ for ${\overrightarrow{E}}_1$ and $\left(-9.54\text{kN}/\text{C}\right)\widehat{i}-\left(9.54\text{kN}/\text{C}\right)\widehat{j}$ for ${\overrightarrow{E}}_2$ in equation (1).

  $\begin{array}{l}{\overrightarrow{E}}_P=\left(1.44\text{kN}/\text{C}\right)\widehat{i}-\left(0.575\text{kN}/\text{C}\right)\widehat{j}+\left(-9.54\text{kN}/\text{C}\right)\widehat{i}-\left(9.54\text{kN}/\text{C}\right)\widehat{j}\\ {\overrightarrow{E}}_P=\left(-8.10\text{kN}/\text{C}\right)\widehat{i}-\left(10.1\text{kN}/\text{C}\right)\widehat{j}\end{array}$

The Magnitude of electric field is:

  $\begin{array}{l}{E}_P=\sqrt{-{\left(8.10\text{kN}/\text{C}\right)}^2-{\left(10.1\text{kN}/\text{C}\right)}^2}\\ E_P=13\text{kN}/\text{C}\end{array}$

Direction of electric field is:

  $\begin{array}{l}\theta ={\tan }^{-1}\left(\frac{-8.10\text{kN}/\text{C}}{-10.1\text{kN}/\text{C}}\right)\\ \theta =230°\end{array}$

Conclusion:

The magnitude and direction of electric field is $13\text{kN}/\text{C}$ and the direction is $230°$ .

(b)

To determine

The magnitude and direction of electric force.

Explanation of Solution

Given:

The magnitude of the first point charge is $-5.0\text{μC}$ .

The position of the first point charge is $\left(4.0\text{m},-\text{2}\text{m}\right)$ .

The magnitude of the second point charge is $1\text{2}\text{μC}$ .

The position of the second point charge is $\left(-1.0\text{m},0\text{m}\right)$ .

Formula used:

Write the expression for the force.

  $\overrightarrow{F}=q\overrightarrow{E}$ ..................(4)

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

Calculation:

Substitute $\left(-8.10\text{kN}/\text{C}\right)\widehat{i}-\left(10.1\text{kN}/\text{C}\right)\widehat{j}$ for $\overrightarrow{E}$ and $1.602\times {10}^{-19}\text{C}$ for $q$ in equation (4).

  $\begin{array}{l}\overrightarrow{F}=1.602\times {10}^{-19}\text{C}\left(\left(-8.10\text{kN}/\text{C}\right)\widehat{i}-\left(10.1\text{kN}/\text{C}\right)\widehat{j}\right)\\ \overrightarrow{F}=\left(1.30\times {10}^{-15}\text{N}\right)\widehat{i}+\left(1.62\times {10}^{-15}\text{N}\right)\widehat{j}\end{array}$

The magnitude of force is:

  $\begin{array}{l}F=\sqrt{{\left(1.30\times {10}^{-15}\text{N}\right)}^2+{\left(1.62\times {10}^{-15}\text{N}\right)}^2}\\ F=2.1\times {10}^{-15}\text{N}\end{array}$

The direction of electric force is:

  $\begin{array}{l}\theta ={\tan }^{-1}\left(\frac{1.62\times {10}^{-15}\text{N}}{1.30\times {10}^{-15}\text{N}}\right)\\ \theta =53°\end{array}$

Conclusion:

The magnitude and direction of force is $2.1\times {10}^{-15}\text{N}$ and $53°$ respectively.