Chapter 21, Problem 33P

To determine

The total electric force on three point charges.

Explanation of Solution

Given:

The magnitude of the first point charge is $-1.0\text{μC}$ .

The magnitude of the second point charge is $2.0\text{μC}$ .

The position of the second point charge particle is $(0,0.10\text{m)}$ .

The magnitude of the third point charge is $4.0\text{μC}$ .

The position of the third point charge is $(0.20\text{m},0)$ .

Formula used:

Write the expression for the net force that acts on $q_1$ .

  ${\overrightarrow{F}}_1={\overrightarrow{F}}_{2,1}+{\overrightarrow{F}}_{3,1}$ ..................(1)

Here, ${\overrightarrow{F}}_1$ is the net force on charge $q_1$ , ${\overrightarrow{F}}_{2,1}$ is the force on first charge due to second charge and ${\overrightarrow{F}}_{3,1}$ is the force on the first charge due to the third charge.

Write the expression for the force on first charge due to second charge.

  ${\overrightarrow{F}}_{2,1}=\frac{k{q}_1{q}_2}{r_{2,1}^3}{\overrightarrow{r}}_{2,1}$ ..................(2)

Here, $r_{2,1}$ is the distance between the first and the second charge.

Write the expression for the force on first charge due to second charge.

  ${\overrightarrow{F}}_{3,1}=\frac{k{q}_1{q}_3}{r_{3,1}^3}{\overrightarrow{r}}_{3,1}$ …….. (3)

Here, $r_{3,1}$ is the distance between the first and the third charge.

Write the expression for the net force that acts on $q_2$ .

  ${\overrightarrow{F}}_2={\overrightarrow{F}}_{1,2}+{\overrightarrow{F}}_{3,2}$ ..................(4)

Here, ${\overrightarrow{F}}_1$ is the net force on charge $q_1$ , ${\overrightarrow{F}}_{2,1}$ is the force on first charge due to second charge and ${\overrightarrow{F}}_{3,1}$ is the force on the first charge due to the third charge.

Write the expression for the force on second charge due to third charge.

  ${\overrightarrow{F}}_{1,2}=\frac{k{q}_1{q}_2}{r_{1,2}^3}{\overrightarrow{r}}_{1,2}$

Here, $r_{1,2}$ is the distance between the first and the second charge.

Write the expression for the force on first charge due to second charge.

  ${\overrightarrow{F}}_{3,2}=\frac{k{q}_1{q}_3}{r_{3,2}^3}{\overrightarrow{r}}_{3,2}$ ..................(5)

Here, $r_{3,2}$ is the distance between the second and the third charge.

Write the expression for the net force that acts on the third charge.

  ${\overrightarrow{F}}_3={\overrightarrow{F}}_{1,3}+{\overrightarrow{F}}_{2,3}$

From Newton’s third law the action and reaction forces are equal that is:

  ${\overrightarrow{F}}_3=-{\overrightarrow{F}}_{3,1}-{\overrightarrow{F}}_{3,2}$ ..................(6)

Here, ${\overrightarrow{F}}_3$ is the net force on the third charge, ${\overrightarrow{F}}_{1,3}$ is the force on the third charge due to first charge and ${\overrightarrow{F}}_{2,3}$ is the force on the third charge due to the second charge.

Calculation:

Substitute $8.99\times {10}^9\text{N}{\text{.m}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $k$ , $0.10\text{m}$ for $r_{1,2}$

  $2.0\text{μC}$ for $q_2$ and $-1.0\text{μC}$ for $q_1$ in equation (2).

  $\begin{array}{l}{\overrightarrow{F}}_{2,1}=\frac{8.99\times {10}^9\text{N}{\text{.m}}^{\text{2}}/{\text{C}}^{\text{2}}\left(2.0\text{μC}\right)-1.0\text{μC}}{{\left(0.10\text{m}\right)}^3}\left(-0.10\text{m}\right)\widehat{j}\\ {\overrightarrow{F}}_{2,1}=\left(1.80\text{N}\right)\widehat{j}\end{array}$

Substitute $8.99\times {10}^9\text{N}{\text{.m}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $k$ , $0.20\text{m}$ for $r_{3,1}$ , $4.0\text{μC}$ for $q_3$ and $-1.0\text{μC}$ for $q_1$ in equation (3).

  $\begin{array}{l}{\overrightarrow{F}}_{3,1}=\frac{8.99\times {10}^9\text{N}{\text{.m}}^{\text{2}}/{\text{C}}^{\text{2}}\left(4.0\text{μC}\right)\left(-1.0\text{μC}\right)}{{\left(0.20\text{m}\right)}^3}\left(-0.20\text{m}\right)\\ {\overrightarrow{F}}_{3,1}=\left(0.899\text{N}\right)\widehat{i}\end{array}$

Substitute $\left(1.80\text{N}\right)\widehat{j}$ for ${\overrightarrow{F}}_{2,1}$ and $\left(0.899\text{N}\right)\widehat{i}$ for ${\overrightarrow{F}}_{3,1}$ in equation (1).

  ${\overrightarrow{F}}_1=\left(1.80\text{N}\right)\widehat{j}+\left(0.899\text{N}\right)\widehat{i}$

Substitute $8.99\times {10}^9\text{N}{\text{.m}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $k$ , $0.224\text{m}$ for $r_{3,1}$ , $4.0\text{μC}$ for $q_3$ and $2.0\text{μC}$ for $q_1$ in equation (5).

  $\begin{array}{l}{\overrightarrow{F}}_{3,2}=\frac{8.99\times {10}^9\text{N}{\text{.m}}^{\text{2}}/{\text{C}}^{\text{2}}\left(4.0\text{μC}\right)-1.0\text{μC}}{{\left(0.224\text{m}\right)}^3}\left(-0.2\text{m}\right)\widehat{i}+\left(1.0\text{m}\right)\widehat{j}\\ {\overrightarrow{F}}_{3,2}=\left(-1.28\text{N}\right)\widehat{i}+\left(0.640\text{N}\right)\widehat{j}\end{array}$

Substitute $\left(-1.28\text{N}\right)\widehat{i}+\left(0.640\text{N}\right)\widehat{j}$ for ${\overrightarrow{F}}_{3,2}$ and $\left(-1.80\text{N}\right)\widehat{j}$ in equation (4).

  $\begin{array}{l}{\overrightarrow{F}}_2=\left(-1.80\text{N}\right)\widehat{j}+\left(-1.28\text{N}\right)\widehat{i}+\left(0.640\text{N}\right)\widehat{j}\\ {\overrightarrow{F}}_2=-\left(1.28\right)\widehat{i}-\left(1.16\text{N}\right)\widehat{j}\end{array}$

Substitute $\left(0.899\text{N}\right)\widehat{i}$ for ${\overrightarrow{F}}_{3,1}$ and $\left(-1.28\text{N}\right)\widehat{i}+\left(0.640\text{N}\right)\widehat{j}$ for ${\overrightarrow{F}}_{3,2}$ in equation (6).

  $\begin{array}{l}{\overrightarrow{F}}_3=-\left(0.899\text{N}\right)\widehat{i}-\left(-1.28\text{N}\right)\widehat{i}+\left(0.640\text{N}\right)\widehat{j}\\ {\overrightarrow{F}}_3=\left(0.381\text{N}\right)\widehat{i}-\left(0.640\text{N}\right)\widehat{j}\end{array}$

Conclusion:

The total electric force on first, second and third charges are $\left(1.80\text{N}\right)\widehat{j}+\left(0.899\text{N}\right)\widehat{i}$ , $-\left(1.28\right)\widehat{i}-\left(1.16\text{N}\right)\widehat{j}$ and $\left(0.381\text{N}\right)\widehat{i}-\left(0.640\text{N}\right)\widehat{j}$ respectively.