Chapter 21, Problem 45P

(a)

To determine

The magnitude and direction of the electric field.

Answer to Problem 45P

The components of the resultant electric field are

  $\begin{array}{l}{E}_x=-3129.05\cos \left({85.49}^o\right)\text{V/m}\\ E_y=-3129.05\sin \left({85.49}^o\right)\text{V/m}\end{array}$

The magnitude of the resultant electric field $3129.05\text{V/m}$ making an angle of $\left({85.49}^o\right)$ with the negative x axis.

Explanation of Solution

Introduction:

Electric field, $\overrightarrow{E}$ at distance r from a charge q is given by the following equation:

  $\overrightarrow{E}=k\frac{q}{r^2}\hat{r}\dots \mathrm{..}\left(1\right)$

  $k$ is the Coulomb constant which is equal to $8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}\cdot {\text{C}}^{\text{-2}}$ .Electric field is a vector quantity, hence, is given by magnitude and the direction. The field lines of a positive charge always direct outwardly away from the charge and the field lines of a negative charge points towards the charge.

Also, two similar charges repel each other and opposite charges attract each other with equal and opposite force.

A charge $q_1=5\text{μC}$ is placed at point A whose co-ordinates are given by (1,3). Another charge $q_2=-4\text{μC}$ is located at point B whose co-ordinates are (2, -2) as shown in Figure 1. The resultant field produced by these two given charges is to be determined at point P whose coordinates are (-3,1).

  

Figure 1: Two charges q1 and q2 are placed at the given co-ordinates, and the electric fieldis to be determined due to thetwo given charges is to be found at point P.

Now at point A, the positive charge $q_1=5\text{μC}$ is placed, hence, the electric field line ${\overrightarrow{E}}_1$ due to this charge at point P will direct away from it as shown with a yellow arrow in Figure 2. Similarly, the field lines ${\overrightarrow{E}}_2$ due to negative charge $q_2=-4\text{μC}$ placed at point B would be towards the charge, therefore, the field lines shown in Figure 2 with pink arrow. The green arrow is the resultant ${\overrightarrow{E}}_R$ of the electric field lines ${\overrightarrow{E}}_1$ and ${\overrightarrow{E}}_2$ . ${\theta }_R$ is the resultant angle. In order to find the magnitude of the eletric field , distanc ebeteen the charges an dthe point P is to be calculated.

The distance between points A and B is

  $\begin{array}{c}AB=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(1-2\right)}^2+{\left(3+2\right)}^2}\\ =\sqrt{26}\text{m}\end{array}$

The distance between points P and B is

  $\begin{array}{c}BP=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(-3-2\right)}^2+{\left(1+2\right)}^2}\\ =\sqrt{34}\text{m}\end{array}$

The distance between points P and A is

  $\begin{array}{c}AP=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(1+3\right)}^2+{\left(3-1\right)}^2}\\ =\sqrt{20}\text{m}\end{array}$

  

Figure 2: The field lines due the two charges at point P and the resultant of the field.

Electric field at distance r from a charge q is given by

  $\overrightarrow{E}=k\frac{q}{r^2}\hat{r}$

  $\begin{array}{c}\left|{\overrightarrow{E}}_1\right|=k\frac{q_1}{{\left(AP\right)}^2}\\ =8.99\times {10}^9\times \frac{5\times {10}^{-6}}{{\left(\sqrt{20}\right)}^2}\\ =2247.5\text{V/m}\\ \left|{\overrightarrow{E}}_2\right|=k\frac{q_2}{{\left(BP\right)}^2}\\ =8.99\times {10}^9\times \frac{4\times {10}^{-6}}{{\left(\sqrt{34}\right)}^2}\\ =1057.64\text{7V/m}\end{array}$

Cosine law is used to find the angle $\theta$ shown in Figure 2 as below:

  $\begin{array}{l}A{P}^2+B{P}^2-2\times AP\times BP\times cos\theta =A{B}^2\\ cos\theta =\frac{A{P}^2+B{P}^2-A{B}^2}{2\times AP\times BP}\end{array}$

Putting the values from equation, we get

  $\begin{array}{l}cos\theta =\frac{20+34-26}{2\times \sqrt{20}\times \sqrt{34}}=\frac{28}{52.15}=0.536\\ \theta ={\cos }^{-1}\left(0.536\right)={57.58}^o\end{array}$

Now, to find the resultant angle, just focus on the filed lines shown in Figure 3. ${\theta }_1$ is the angle between the two electric fields ${\overrightarrow{E}}_1$ and ${\overrightarrow{E}}_2$ shown in Figure 3 and can be calculated as below:

  $\begin{array}{c}{\theta }_1{\text{=180}}^{\text{o}}-{\theta }^{\text{o}}\\ {\text{=180}}^{\text{o}}\text{-57}{\text{.58}}^{\text{o}}\\ \text{=122}{\text{.42}}^{\text{o}}\end{array}$

  

Figure 3: The resultant field ${\text{<mover accent="true"> <mi>E</mi> <mo>→</mo></mover>}}_{\text{R}}$ and resultant field angle ${\text{θ}}_{\text{R}}$

Now the using the parallelogram theorem, we can find the resultant of the field ${\overrightarrow{E}}_1$

and ${\overrightarrow{E}}_2$ as shown in Figure 3.

  $\begin{array}{l}\left|E_R\right|=E_1^2+E_2^2+2E_1{E}_2\cos {\theta }_1\\ \left|E_R\right|=\sqrt{{\left(2247.5\right)}^2+{\left(1057.647\right)}^2+2\times 2247.5\times 1057.647\times \left(\text{-0}\text{.53612149}\right)}\\ \left|E_R\right|=1902.94\text{V/m}\end{array}$

The resultant angle is calculated as below:

  $\begin{array}{c}\tan {\theta }_R=\frac{E_2\sin {\theta }_1}{E_1+E_2\cos {\theta }_1}\\ =\frac{1057.647\sin \left({122.42}^o\right)}{2247.5+1057.647\times \cos \left({122.42}^o\right)}\\ =\frac{892.80}{1680.47}\\ =0.5312\\ {\theta }_R={27.98}^o\end{array}$

In order to find the angle of resultant electric field, trigonometry is used as shown in Figure 4:

  

Figure 4: The angle of resultant electric field with the x axis

  ${\theta }_x$ can be calculated by finding the slope of the line BP using the coordinate values which can be calculate as

  $\begin{array}{l}tan{\theta }_x=\frac{y_2-y_1}{x_2-x_1}=-\frac{3}{5}\\ {\theta }_x={30}^o\\ {\theta }_x+{\theta }_1+{\theta }_E={180}^o\\ {30}^o+{122.42}^o+{\theta }_E={180}^o\\ {\theta }_E={27.58}^o\end{array}$

The angle of the ${\overrightarrow{E}}_R$ with the negative x axis is ${\theta }_E+{\theta }_R={27.58}^o+{27.98}^o={55.56}^o$

Thus, the

  $\begin{array}{l}{E}_x=-E_R\cos \left({\theta }_E+{\theta }_R\right)=-1902.94\cos \left({55.56}^o\right)\text{V/m}\\ E_y=-E_R\sin \left({\theta }_E+{\theta }_R\right)=-1902.94\sin \left({55.56}^o\right)\text{V/m}\end{array}$

Conclusion:

The components of the resultant electric field is

  $\begin{array}{l}{E}_x=-E_R\cos \left({\theta }_E+{\theta }_R\right)=-1902.94\cos \left({55.56}^o\right)\text{V/m}\\ E_y=-E_R\sin \left({\theta }_E+{\theta }_R\right)=-1902.94\sin \left({55.56}^o\right)\text{V/m}\end{array}$

The magnitude of the resultant electric field $\text{1902}\text{.94V/m}$ making an angle of $\left({55.56}^o\right)$ with the negative x axis.

(b)

To determine

The magnitude and direction of the force of the proton.

Answer to Problem 45P

The components of the resultant force is :

  $\begin{array}{l}{F}_x=-3.04\times {10}^{-16}\cos \left({55.56}^o\right)\text{N}\\ F_y=-3.04\times {10}^{-16}\sin \left({55.56}^o\right)\text{N}\end{array}$

The magnitude of the resultant force is $3.04\times {10}^{-16}\text{N}$ making an angle of $\left({55.56}^o\right)$ with the negative x axis.

Explanation of Solution

Introduction:

The electric force between two charges $q_1$ and $q_2$ placed at distance r exert force on each other is given as below:

  $\overrightarrow{F}=k\frac{q_1\times q_2}{r^2}\hat{r}$

  $k$ Coulomb’s constant which equals to $8.99\times {10}^9{\text{Nm}}^{\text{2}}{\text{C}}^{\text{-2}}$ .

Also, two similar charges repel each other and opposite charges attract each other with equal and opposite force.

A charge $q_1=5\mu C$ is placed at point A whose co-ordinates are given by (1,3). Another charge $q_2=-4\mu C$ is located at point B whose co-ordinates are (2, -2) as shown in Figure 5.A proton is placed at point Pwhose coordinates are (-3,1). There is arepulsiveforce on the proton due to charge $q_1=5\mu C$ as both are positive in nature. The magnitude of the force is given as

  $\begin{array}{c}{\text{F}}_{\text{1p}}\text{=k}\frac{{\text{q}}_{\text{1}}{\text{×q}}_{\text{p}}}{{\text{AP}}^{\text{2}}}\\ \text{=8}{\text{.99×10}}^{\text{9}}\text{×}\frac{{\text{5×10}}^{\text{-6}}\text{×1}{\text{.6×10}}^{\text{-19}}}{{\left(\sqrt{\text{20}}\right)}^{\text{2}}}\\ \text{=3}{\text{.596×10}}^{\text{-16}}\text{N}\end{array}$

There is an attractive force on the proton due to charge $q_2=-4\text{μC}$ as both are opposite charge in nature. The magnitude of the force is given as

  $\begin{array}{c}{F}_{2p}=k\frac{q_2\times q_p}{B{P}^2}\\ =-8.99\times {10}^9\times \frac{4\times {10}^{-6}\times 1.6\times {10}^{-19}}{{\left(\sqrt{34}\right)}^2}\\ =-1.699\times {10}^{-16}\text{N}\end{array}$

  

Figure 5: Forces acting on the proton placed at point P

The resultant force acting on the proton will be:

  $\left|F\right|=F_{p1}^2+E_{p2}^2+2F_{p1}{F}_{p2}\cos {\theta }_1$

  ${\theta }_1$ is already calculated in the previous section which ${122.42}^o$ .

  $\left|F\right|=\sqrt{{\left(3.596\times {10}^{-16}\right)}^2+{\left(1.699\times {10}^{-16}\right)}^2+2\times 1.699\times {10}^{-16}\times 3.596\times {10}^{-16}\times \cos {122.42}^o}$

  $\left|F\right|=\sqrt{9.26}\times {10}^{-16}=3.04\times {10}^{-16}\text{N}$

The direction of the force will be the same as of the field calculated in the previous section.

Conclusion:

The components of the resultant force is :

  $\begin{array}{l}{F}_x=-3.04\times {10}^{-16}\cos \left({55.56}^o\right)\text{N}\\ F_y=-3.04\times {10}^{-16}\sin \left({55.56}^o\right)\text{N}\end{array}$

The magnitude of the resultant force is $3.04\times {10}^{-16}\text{N}$ making an angle of $\left({55.56}^o\right)$ with the negative x axis.