Chapter 21, Problem 76P

(a)

To determine

The magnitude and the direction of the force exerted.

Explanation of Solution

Given:

Four point charges placed at corners of a square whose edge length is L.

Formula used:

Write the expression for the net force acting on ${\overrightarrow{F}}_1$ .

  ${\overrightarrow{F}}_1={\overrightarrow{F}}_{2,1}+{\overrightarrow{F}}_{3,1}+{\overrightarrow{F}}_{4,1}$   ........ (1)

Here, ${\overrightarrow{F}}_1$ is the net force on charge $q_1$ , ${\overrightarrow{F}}_{2,1}$ is the force on charge $q_1$ due to the charge $q_2$ , ${\overrightarrow{F}}_{3,1}$ is the charge on charge $q_1$ due to the charge $q_3$ , ${\overrightarrow{F}}_{4,1}$ is the force on charge $q_1$ due o the charge $q_4$ .

Write the expression for the ${\overrightarrow{F}}_{2,1}$ that is the force on charge $q_1$ due to the charge $q_2$ .

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

Here, $k$ is the constant.

Write the expression for the ${\overrightarrow{F}}_{3,1}$ that is the force on charge $q_1$ due to the charge $q_3$ .

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

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

The magnitude of charge $q_4$ is three times that of the other charges that is:

  ${\overrightarrow{F}}_{4,1}=3\frac{k{q}_1{q}_2}{{\left(r_{4,1}\right)}^2}{\widehat{r}}_{4,1}$   ........ (4)

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

Substitute $-q$ for $q_1$ , $q$ for $q_1$ and $L$ for $r_{2,1}$ in equation (2).

  $\begin{array}{l}{\overrightarrow{F}}_{2,1}=\frac{k\left(-q\right)q}{L^3}\left(-L\widehat{j}\right)\\ {\overrightarrow{F}}_{2,1}=\frac{k{q}^2}{L^2}\widehat{j}\end{array}$

Substitute $q$ for $q_1$ , $q$ for $q_2$ and $L$ for $r_{3,1}$ in equation (3).

  $\begin{array}{l}{\overrightarrow{F}}_{3,1}=\frac{k{q}^2}{2^{\frac{3}{2}}{L}^3}\left(-L\widehat{i}-L\widehat{j}\right)\\ {\overrightarrow{F}}_{3,1}=\frac{-k{q}^2}{2^{\frac{3}{2}}{L}^2}\left(\widehat{i}+\widehat{j}\right)\end{array}$

Substitute $-q$ for $q_1$ , $q$ for $q_1$ and $L$ for $r_{4,1}$ in equation (4).

  $\begin{array}{l}{\overrightarrow{F}}_{4,1}=\frac{k\left(-q\right)q}{L^3}\left(-L\widehat{i}\right)\\ {\overrightarrow{F}}_{4,1}=\frac{k{q}^2}{L^2}\left(\widehat{i}\right)\end{array}$

Calculation:

Substitute $\frac{k{q}^2}{L^2}\widehat{j}$ for ${\overrightarrow{F}}_{2,1}$ , $\frac{-k{q}^2}{2^{\frac{3}{2}}{L}^2}\left(\widehat{i}+\widehat{j}\right)$ for ${\overrightarrow{F}}_{3,1}$ , $\frac{k{q}^2}{L^2}\left(\widehat{i}\right)$ for ${\overrightarrow{F}}_{4,1}$ in equation (1).

  $\begin{array}{l}{\overrightarrow{F}}_1=\frac{k{q}^2}{L^2}\widehat{j}+\frac{-k{q}^2}{2^{\frac{3}{2}}{L}^2}\left(\widehat{i}+\widehat{j}\right)+\frac{k{q}^2}{L^2}\left(\widehat{i}\right)\\ {\overrightarrow{F}}_1=\frac{k{q}^2}{L^2}\left(1-\frac{1}{2\sqrt{2}}\right)\left(\widehat{i}+\widehat{j}\right)\end{array}$

Conclusion:

Thus, the magnitude and the direction of the force is $\frac{k{q}^2}{L^2}\left(1-\frac{1}{2\sqrt{2}}\right)\left(\widehat{i}+\widehat{j}\right)$ .

(b)

To determine

The magnitude of the electric field.

Explanation of Solution

Formula used:

Write the expression for the electric field at $E_P$ .

  ${\overrightarrow{E}}_P={\overrightarrow{E}}_1+{\overrightarrow{E}}_2+{\overrightarrow{E}}_3+{\overrightarrow{E}}_4$   ......... (5)

Here, ${\overrightarrow{E}}_P$ is the electric field at the point where the electric field is calculated, ${\overrightarrow{E}}_1$ is the electric field due to charge $q_1$ , ${\overrightarrow{E}}_2$ is the electric field due to charge $q_2$ , ${\overrightarrow{E}}_3$ is the electric field due to charge $q_3$ and ${\overrightarrow{E}}_4$ is the electric field due to charge $q_4$ .

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

  ${\overrightarrow{E}}_1=\frac{kq}{r_{1,p}^2}\left(\frac{L}{2}\widehat{j}\right)$   ......... (6)

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

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

  ${\overrightarrow{E}}_2=\frac{k\left(-q\right)}{r_{2,p}^2}\left(-\frac{L}{2}\widehat{j}\right)$   ......... (7)

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

Write the expression for the electric field due to charge $q_3$ .

  ${\overrightarrow{E}}_3=\frac{k{q}_3}{r_{3,p}^2}\left(-L\widehat{i}-\frac{L}{2}\widehat{j}\right)$   ........ (8)

Here, $r_{3.p}$ is the distance between the third charge and point $p$ and $k$ is constant.

Write the expression for the electric field due to charge $q_3$ .

  ${\overrightarrow{E}}_{4.p}=\frac{k\left(-q\right)}{r_{4,p}^2}\left(L\widehat{i}-\frac{L}{2}\widehat{j}\right)$   ........ (9)

Here, $r_{4.p}$ is the distance between the fourth charge and point $p$ and $k$ is constant.

Substitute $\frac{L}{2}$ for $r_{1,p}$ in equation (6).

  $\begin{array}{l}{\overrightarrow{E}}_1=\frac{kq}{{\left(\frac{L}{2}\right)}^2}\left(\frac{L}{2}\widehat{j}\right)\\ {\overrightarrow{E}}_1=\frac{4kq}{L^2}\widehat{j}\end{array}$

Substitute $\frac{L}{2}$ for $r_{2,p}$ in equation (7).

  $\begin{array}{l}{\overrightarrow{E}}_2=\frac{k\left(-q\right)}{{\left(\frac{L}{2}\right)}^2}\left(-\frac{L}{2}\widehat{j}\right)\\ {\overrightarrow{E}}_2=\frac{4kq}{L^2}\widehat{j}\end{array}$

Substitute $L$ for $r_{3,p}$ in equation (8).

  ${\overrightarrow{E}}_3=\frac{k{q}_3}{5^{\frac{3}{2}}{L}^2}\left(-\widehat{i}-\frac{1}{2}\widehat{j}\right)$

Substitute $L$ for $r_{4,p}$ in equation (9).

  ${\overrightarrow{E}}_4=\frac{k{q}_4}{5^{\frac{3}{2}}{L}^2}\left(\widehat{i}-\frac{1}{2}\widehat{j}\right)$

Calculation:

Substitute $\frac{4kq}{L^2}\widehat{j}$ for ${\overrightarrow{E}}_1$ , $\frac{4kq}{L^2}\widehat{j}$ for ${\overrightarrow{E}}_2$ , $\frac{k{q}_3}{5^{\frac{3}{2}}{L}^2}\left(-\widehat{i}-\frac{1}{2}\widehat{j}\right)$ for ${\overrightarrow{E}}_3$ and $\frac{k{q}_4}{5^{\frac{3}{2}}{L}^2}\left(\widehat{i}-\frac{1}{2}\widehat{j}\right)$ for ${\overrightarrow{E}}_4$ in equation (5).

  $\begin{array}{l}{\overrightarrow{E}}_P=\frac{4kq}{L^2}\widehat{j}+\frac{4kq}{L^2}\widehat{j}+\frac{k{q}_3}{5^{\frac{3}{2}}{L}^2}\left(-\widehat{i}-\frac{1}{2}\widehat{j}\right)+\frac{k{q}_4}{5^{\frac{3}{2}}{L}^2}\left(\widehat{i}-\frac{1}{2}\widehat{j}\right)\\ {\overrightarrow{E}}_P=\frac{8kq}{L^2}\left(1+\frac{\sqrt{5}}{25}\right)\widehat{j}\end{array}$

Conclusion:

Thus, the magnitude and the direction of the electric field is $\frac{8kq}{L^2}\left(1+\frac{\sqrt{5}}{25}\right)\widehat{j}$ .