Chapter 16, Problem 58QAP

To determine

(a)

The electric field at point (0, 3) cm

Answer to Problem 58QAP

Net electric field is $5.09\times {10}^5\text{N/C}$ at an angle $26.01°$ with the x -axis.

Explanation of Solution

Given:

Two charges are placed on the x -axis:

  $q_1=+1.0\times {10}^{-7}\text{C}$ at $x_1=0.05\text{ m}$ and

  $q_2=2.00\times {10}^{-2}\text{C}$ at $x_2=0.08\text{ m}$.

Formula used:

Electric field at distance r from the charge Q.

  $E=\frac{k\left|Q\right|}{r^2}$

Calculation:

Point P is at x =0, y = 0.03 m

  

  $\begin{array}{l}{\theta }_1={\tan }^{-1}\left(\frac{0.03}{0.05}\right)={30.96}^o\\ {\theta }_2={\tan }^{-1}\left(\frac{0.03}{0.08}\right)={20.55}^o\\ r^2=x^2+y^2\\ r_1^2={0.05}^2+{0.03}^2=3.4\times {10}^{-3}{m}^2\\ r_2^2={0.08}^2+{0.03}^2=7.3\times {10}^{-3}{m}^2\end{array}$

The x -component of electric field is:

  $\begin{array}{l}{E}_x=E_{1x}+E_{2x}=\frac{k{q}_1}{r_1^2}\cos \left({\theta }_1\right)+\frac{k{q}_2}{r_2^2}\cos \left({\theta }_2\right)=k\left[\frac{q_1}{r_1^2}\cos \left({\theta }_1\right)+\frac{q_2}{r_2^2}\cos \left({\theta }_2\right)\right]\\ =\left(8.99\times {10}^9\frac{\text{N}\cdot {\text{m}}^2}{{\text{C}}^2}\right)[\frac{\left(1.0\times {10}^{-7}\text{C}\right)}{\left(3.4\times {10}^{-3}{m}^2\right)}\cos \left({30.96}^o\right)\\ +\frac{\left(2.0\times {10}^{-7}\text{C}\right)}{\left(7.3\times {10}^{-3}{m}^2\right)}\cos \left({20.55}^o\right)]=4.573\times {10}^5\frac{\text{N}}{\text{C}}\\ =4.57\times {10}^5\frac{\text{N}}{\text{C}}(\text{ rounded to }3\text{ significant figures })\end{array}$

The y -component of electric field is:

  $\begin{array}{l}{E}_y=E_{1y}+E_{2y}=\frac{k{q}_1}{r_1^2}\sin \left({\theta }_1\right)+\frac{k{q}_2}{r_2^2}\sin \left({\theta }_2\right)=k\left[\frac{q_1}{r_1^2}\sin \left({\theta }_1\right)+\frac{q_2}{r_2^2}\sin \left({\theta }_2\right)\right]\\ =\left(8.99\times {10}^9\frac{\text{N}\cdot {\text{m}}^2}{{\text{C}}^2}\right)[\frac{\left(1.0\times {10}^{-7}\text{C}\right)}{\left(3.4\times {10}^{-3}{m}^2\right)}\sin \left(30.96\right)\\ +\frac{\left(2.0\times {10}^{-7}\text{C}\right)}{\left(7.3\times {10}^{-3}{m}^2\right)}\sin \left(20.55\right)]=2.2252\times {10}^5\frac{\text{N}}{\text{C}}\\ =2.23\times {10}^5\frac{\text{N}}{\text{C}}(\text{rounded}\text{to}\text{3}\text{significant}\text{figures})\end{array}$

Net electric field,

  $\begin{array}{l}E=\sqrt{E_x+E_y}\\ E=\sqrt{{\left(4.57\times {10}^5\frac{N}{C}\right)}^2+{\left(2.23\times {10}^5\frac{N}{C}\right)}^2}\\ E=5.09\times {10}^5\frac{N}{C}\end{array}$

At the angle,

  $\begin{array}{l}\alpha ={\tan }^{-1}\left(\frac{E_y}{E_x}\right)={\tan }^{-1}\left(\frac{\left(2.23\times {10}^5\frac{\text{N}}{\text{C}}\right)}{\left(4.57\times {10}^5\frac{\text{N}}{\text{C}}\right)}\right)\\ \alpha =26.01{}^o\end{array}$

Conclusion:

Net electric field is $5.09\times {10}^5\text{N/C}$ at an angle $26.01°$ with the x -axis.

To determine

(b)

Force acting on the electron at point (0, 3) cm

Answer to Problem 58QAP

Net force acting on the electron is $-8.15\times {10}^{-14}\text{N}$

Direction of force on the electron is toward the positive charges.

Explanation of Solution

Given:

Electric field, $E=5.09\times {10}^5\text{N/C}$

Charge on electron, $q=-1.602\times {10}^{-19}\text{C}$

Formula used:

  $\overrightarrow{F}=q\overrightarrow{E}$

Calculation:

  $\begin{array}{l}\overrightarrow{F}=q\overrightarrow{E}\\ \overrightarrow{F}=qE=\left(-1.602\times {10}^{-19}\text{C}\right)\left(5.09\times {10}^5\frac{\text{N}}{\text{C}}\right)=-8.15\times {10}^{-14}\text{N}\end{array}$

Conclusion:

Net force acting on electron is $-8.15\times {10}^{-14}\text{N}$, negative sign indicates that direction of force on the electron is opposite to the direction of electric field.