Chapter 21, Problem 30P

To determine

The force exerted by the three point charges on the fourth charge.

Explanation of Solution

Given: The magnitude of the three point charges is $3.00\text{nC}$ .

The length of the side of the square is $5.00\text{cm}$ .

Formula used:

Write the expression for the coulomb law.

  $F=\frac{k{q}_1{q}_2}{r^2}$

Here, $F$ is the force, $k$ is the constant, $q_1$ is the point charge, $q_2$ is the second point charge and $r$ is the distance between the two point charges.

The representation of the four charges at the vertices of the square is shown below:

  

Here, ${\overrightarrow{F}}_{1,4}$ is the force on the fourth charge due to first charge, ${\overrightarrow{F}}_{3,4}$ is the force on the fourth charge due to third charge, ${\overrightarrow{F}}_{2,4}$ is the force fourth charge due to second charge.

Write the expression for the net force acting on the fourth charge.

  ${\overrightarrow{F}}_4={\overrightarrow{F}}_{1,4}+{\overrightarrow{F}}_{2,4}+{\overrightarrow{F}}_{3,4}$   ...... (1)

Write the expression for the force exerted on fourth charge due to first charge.

  ${\overrightarrow{F}}_{1,4}=\frac{k{q}_1{q}_4}{r_{1,4}^2}\widehat{j}$   ...... (2)

Here, $r_{1,4}$ is the distance between the first charge and the fourth charge.

Write the expression for the force exerted by the second charge on the fourth charge.

  ${\overrightarrow{F}}_{2,4}=\frac{k{q}_2{q}_4}{r_{2,4}^2}\widehat{i}$   ...... (3)

Here, $r_{2,4}$ is the distance between the second point charge and the fourth point charge.

Write the expression for the force exerted by the third charge on the fourth charge.

  ${\overrightarrow{F}}_{3,4}=\frac{k{q}_3{q}_4}{r_{3,4}^2}{\widehat{r}}_{3,4}$   ...... (4)

Here, ${\widehat{r}}_{3,4}$ is the unit vector that points from $q_3$ to $q_4$ , $r_{3,4}$ is the distance between the third charge and the fourth charge.

Write the expression for ${\overrightarrow{r}}_{3,4}$ in terms of distance between the third and the first charge and between the first and the fourth charge.

  ${\overrightarrow{r}}_{3,4}={\overrightarrow{r}}_{3,1}+{\overrightarrow{r}}_{1,4}$   ...... (5)

Write the expression for the unit vector.

  ${\widehat{r}}_{3,4}=\frac{{\overrightarrow{r}}_{3,4}}{\left|{\overrightarrow{r}}_{3,4}\right|}$

  $$   ...... (6)

Here, $\left|{\overrightarrow{r}}_{3,4}\right|$ is the magnitude of position vector.

Calculation:

Substitute $8.988\times {10}^9{\text{Nm}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $k$ , $3.00\text{nC}$ for $q_2$ , $3.00\text{nC}$ for $q_4$ , $0.05\text{m}$ for $r_{1,4}$ in equation (2).

  $\begin{array}{l}{\overrightarrow{F}}_{1,4}=\frac{\left(8.988\times {10}^9{\text{Nm}}^{\text{2}}/{\text{C}}^{\text{2}}\right){\left[3.00\text{nC}\right]}^2}{{\left(0.05\text{m}\right)}^2}\widehat{j}\\ {\overrightarrow{F}}_{1,4}=\left(3.23\times {10}^{-5}\text{N}\right)\widehat{j}\end{array}$

Substitute $8.988\times {10}^9{\text{Nm}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $k$ , $3.00\text{nC}$ for $q_1$ , $3.00\text{nC}$ for $q_4$ , $0.05\text{m}$ for $r_{1,4}$ in equation (3).

  $\begin{array}{l}{\overrightarrow{F}}_{2,4}=\frac{\left(8.988\times {10}^9{\text{Nm}}^{\text{2}}/{\text{C}}^{\text{2}}\right){\left[3.00\text{nC}\right]}^2}{{\left(0.05\text{m}\right)}^2}\widehat{i}\\ {\overrightarrow{F}}_{2,4}=\left(3.23\times {10}^{-5}\text{N}\right)\widehat{i}\end{array}$

Substitute $\left(0.05\text{m}\right)\widehat{i}$ for ${\overrightarrow{r}}_{3,1}$ and $\left(0.05\text{m}\right)\widehat{j}$ for ${\overrightarrow{r}}_{1,4}$ in equation (5).

  ${\overrightarrow{r}}_{3,4}=\left(0.05\text{m}\right)\widehat{i}+\left(0.05\text{m}\right)\widehat{j}$

Substitute $\left(0.05\text{m}\right)\widehat{i}+\left(0.05\text{m}\right)\widehat{j}$ for ${\overrightarrow{r}}_{3,4}$ in equation (6).

  $\begin{array}{l}{\widehat{r}}_{3,4}=\frac{\left(0.05\text{m}\right)\widehat{i}+\left(0.05\text{m}\right)\widehat{j}}{\left|{\left(0.05\text{m}\right)}^2\text{+}{\left(0.05\text{m}\right)}^2\right|}\\ {\widehat{r}}_{3,4}=0.707\widehat{i}+0.707\widehat{j}\end{array}$

Substitute $8.988\times {10}^9{\text{Nm}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $k$ , $-3.00\text{nC}$ for $q_3$ , $3.00\text{nC}$ for $q_4$ , $0.707\widehat{i}+0.707\widehat{j}$ for ${\widehat{r}}_{3,4}$

  $0.05\sqrt{2}\text{m}$ for $r_{3,4}$ in equation (4).

  $\begin{array}{l}{\overrightarrow{F}}_{3,4}=\frac{\left(8.988\times {10}^9{\text{Nm}}^{\text{2}}/{\text{C}}^{\text{2}}\right)-3.00\text{nC}\left(3.00\text{nC}\right)}{{\left(0.05\sqrt{2}\text{m}\right)}^2}0.707\widehat{i}+0.707\widehat{j}\\ {\overrightarrow{F}}_{3,4}=-\left(1.14\times {10}^{-5}\text{N}\right)\widehat{i}-\left(1.14\times {10}^{-5}\text{N}\right)\widehat{j}\end{array}$

Substitute $\left(3.23\times {10}^{-5}\text{N}\right)\widehat{j}$ for ${\overrightarrow{F}}_{1,4}$ , $\left(3.23\times {10}^{-5}\text{N}\right)\widehat{i}$ for ${\overrightarrow{F}}_{2,4}$ and $-\left(1.14\times {10}^{-5}\text{N}\right)\widehat{i}-\left(1.14\times {10}^{-5}\text{N}\right)\widehat{j}$ for ${\overrightarrow{F}}_{3,4}$ in equation (1).

  $\begin{array}{l}{\overrightarrow{F}}_4=\left(3.23\times {10}^{-5}\text{N}\right)\widehat{j}+\left(3.23\times {10}^{-5}\text{N}\right)\widehat{i}-\left(1.14\times {10}^{-5}\text{N}\right)\widehat{i}-\left(1.14\times {10}^{-5}\text{N}\right)\widehat{j}\\ {\overrightarrow{F}}_4=\left(2.5\times {10}^{-5}\text{N}\right)\widehat{i}+\left(2.5\times {10}^{-5}\text{N}\right)\widehat{j}\end{array}$

Conclusion:

The force exerted by the three point charges on the fourth charge is $\left(2.5\times {10}^{-5}\text{N}\right)\widehat{i}+\left(2.5\times {10}^{-5}\text{N}\right)\widehat{j}$ .