Chapter 16, Problem 13P

To determine

To calculate: the magnitude and direction of the net force on each particle

Explanation of Solution

Given:

Three positive particles of equal charge $+11.0\mathrm{\mu }\text{C}$ are located at the corners of an equilateral triangle of side $15.0\text{cm}$

Formula used:

According to coulomb's law the magnitude of force $\mathrm{F}$ of attraction or repulsion between two charged particles is,

  $\mathrm{F}=\frac{\mathrm{k}{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{r}}^2}$

Here, $\mathrm{k}$ is the constant, ${\mathrm{q}}_1$ and ${\mathrm{q}}_2$ are the charges, and r is the distance between ${\mathrm{q}}_1$ and ${\mathrm{q}}_2$

The direction of force is along the line joining the two charged particles.

Calculation:

The three charges are same in magnitude, so the magnitude of force between any two particles is equal but differ in direction.

The magnitude of force between any two charged particles is,

  $\mathrm{F}=\frac{\mathrm{k}{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{r}}^2}$

Substitute $9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\text{ for }\mathrm{k},15.0\text{cm for }\mathrm{r},\text{and}+11.0\mathrm{\mu }\text{C for }{\mathrm{q}}_1\text{ and }{\mathrm{q}}_2$

  $\begin{array}{l}\mathrm{F}=\frac{\left(9\times {10}^{\circ }\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)(+11.0\mathrm{\mu }\text{C})(+11.0\mathrm{\mu }\text{C})}{{(15.0\text{cm})}^2}\\ =\frac{\left(9\times {10}^{\circ }\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)(+11.0\mathrm{\mu }\text{C})\left(\frac{{10}^{-6}\text{C}}{1\mathrm{\mu }\text{C}}\right)(+11.0\mathrm{\mu }\text{C})\left(\frac{{10}^{-6}\text{C}}{1\mathrm{\mu }\text{C}}\right)}{{\left(15.0\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^2}\\ \mathrm{F}=48.4\mathrm{N}\end{array}$

Therefore, the force between any two charged particles is $\mathrm{F}=48.4\mathrm{N}$

  

The forces on the $3^{\text{rd}}$ charged particle is shown in the above diagram

From figure, the force acting on the $3^{\text{rd}}$ particle due to $1^{\text{st}}$ particle is,

  ${\overrightarrow{\mathrm{F}}}_{31}={\mathrm{F}}_{31}\cos {60}^{\circ }\text{i}+{\mathrm{F}}_{31}\sin {60}^{\circ }\text{j}$

Here ${\mathrm{F}}_{31}$ is the force on $3^{\text{rd}}$ particle due to $1^{\text{st}}$ particle.

From figure, the force acting on the $3^{\text{rd}}$ particle due to $2^{\text{nd}}$ particle is,

  ${\overrightarrow{\mathrm{F}}}_{32}={\mathrm{F}}_{32}\cos {60}^{\circ }(-\text{i})+{\mathrm{F}}_{32}\sin {60}^{\circ }\text{j}$

Here, ${\mathrm{F}}_{31}$ is the force on $3^{\text{rd}}$ particle due to $1^{\text{st}}$ particle

The net force acting on the $3^{\text{rd}}$ particle is

  ${\overrightarrow{\mathrm{F}}}_3={\overrightarrow{\mathrm{F}}}_{31}+{\overrightarrow{\mathrm{F}}}_{32}$

  $=\left({\mathrm{F}}_{31}\cos {60}^{\circ }\text{i}+{\mathrm{F}}_{31}\sin {60}^{\circ }\text{j}\right)+\left({\mathrm{F}}_{32}\cos {60}^{\circ }(-\text{i})+{\mathrm{F}}_{32}\sin {60}^{\circ }\text{j}\right)$

  ${\overrightarrow{\mathrm{F}}}_3={\mathrm{F}}_{31}\sin {60}^{\circ }\text{j}+{\mathrm{F}}_{32}\sin {60}^{\circ }\text{j}$

Substitute $48.4\text{N}$ for ${\mathrm{F}}_{31}$ and ${\mathrm{F}}_{32}$

  ${\overrightarrow{\mathrm{F}}}_3=(48.4\text{N})\sin {60}^{\circ }\text{j}+(48.4\text{N})\sin {60}^{\circ }\text{j}$

  ${\overrightarrow{\mathrm{F}}}_3=(83.8\text{N})\text{<mover accent="true"><mn>j</mn><mo>^</mo></mover>}$

The magnitude of force acting on the $3^{\text{td}}$ charged particle is,

  ${\mathrm{F}}_3=\sqrt{{\mathrm{F}}_{3\mathrm{x}}^2+{\mathrm{F}}_{3\mathrm{y}}^2}$

Here, ${\mathrm{F}}_{\text{3x}}$ and ${\mathrm{F}}_{\text{3y}}$ are the $\text{X}$ and $\text{Y}$ components of force ${\overrightarrow{\mathrm{F}}}_3$

  $\text{Substitute }0\text{ N for }{\mathrm{F}}_{3\mathrm{x}}\text{ and }83.8\text{N for }{\mathrm{F}}_{3\mathrm{y}}$

  ${\mathrm{F}}_3=\sqrt{{(0\text{N})}^2+{(83.8\text{N})}^2}$

  ${\mathrm{F}}_3=83.8\text{N}$

Therefore, the magnitude of force on $3^{\text{rd}}$ charged particle is $83.8\text{N}$

Therefore, the direction of net force on $1^{\text{st}}$ particle is along positive Y-axis i. $\mathrm{e},$ at an angle of ${90}^{\circ }$ above the positive $\mathrm{x}$ axis

The force acting on charged particles 1 and 2 and their resolution is shown in the figure given below:

  

From figure, the force acting on the $1^{\text{st}}$ particle due to $3^{\text{rd}}$ particle is,

  ${\overrightarrow{\mathrm{F}}}_{13}={\mathrm{F}}_{13}\cos {60}^{\circ }(-\widehat{\mathrm{i}})+{\mathrm{F}}_{13}\sin {60}^{\circ }(-\widehat{\mathrm{j}})$

Here, ${\mathrm{F}}_{31}$ is the force on $1^{\text{st}}$ particle due to $3^{\text{rd}}$ particle.

From figure, the force acting on the $1^{\text{st}}$ particle due to $2^{\text{nd}}$ particle is,

  ${\overrightarrow{\mathrm{F}}}_{12}={\mathrm{F}}_{12}\cos {60}^{\circ }(-\widehat{\mathrm{i}})$

Here, ${\mathrm{F}}_{12}$ is the force on $1^{\text{st}}$ particle due to $2^{\text{nd}}$ particle.

The net force acting on the $1^{\text{st}}$ particle is,

  $\begin{array}{l}{\overrightarrow{\mathrm{F}}}_1={\overrightarrow{\mathrm{F}}}_{13}+{\overrightarrow{\mathrm{F}}}_{12}\\ =\left({\mathrm{F}}_{12}\cos {60}^{\circ }(-\widehat{\mathrm{i}})\right)+\left({\mathrm{F}}_{13}\cos {60}^{\circ }(-\widehat{\mathrm{i}})+{\mathrm{F}}_{13}\sin {60}^{\circ }(-\widehat{\mathrm{j}})\right)\end{array}$

Substitute $48.4\text{N for }{\mathrm{F}}_{12}\text{ and }{\mathrm{F}}_{13}$

  $\begin{array}{l}{\overrightarrow{\mathrm{F}}}_1=\left((48.4\text{N})\cos {60}^{\circ }(-\text{<mover accent="true"><mn> i</mn><mo>^</mo></mover>})\right)+\left((48.4\text{N})\cos {60}^{\circ }(-\text{<mover accent="true"><mn> i</mn><mo>^</mo></mover>})+(48.4\text{N})\sin {60}^{\circ }(-\text{<mover accent="true"><mn> j</mn><mo>^</mo></mover>})\right)\hfill \\ {\overrightarrow{\mathrm{F}}}_1=(48.4\text{N})(-\text{<mover accent="true"><mn>i</mn><mo>^</mo></mover>})+(41.9\text{N})(-\text{<mover accent="true"><mn>j</mn><mo>^</mo></mover>})\hfill \end{array}$

The magnitude of force acting on the $1^{\text{st}}$ charged particle is,

  ${\mathrm{F}}_1=\sqrt{{\mathrm{F}}_{1\mathrm{x}}^2+{\mathrm{F}}_{1\mathrm{y}}^2}$

Here, ${\mathrm{F}}_{1\mathrm{x}}\text{ and }{\mathrm{F}}_{1\mathrm{y}}$ are the $\mathrm{X}$ and $\mathrm{Y}$ components of force ${\overrightarrow{\mathrm{F}}}_1$

Substitute $48.4\text{N}$ for ${\mathrm{F}}_{1\mathrm{x}}$ and $41.9\text{N}$ for ${\mathrm{F}}_{1\mathrm{y}}$

  $\begin{array}{l}{\mathrm{F}}_1=\sqrt{{(48.4\text{N})}^2+{(41.9\text{N})}^2}\hfill \\ {\mathrm{F}}_1=64\text{N}\hfill \end{array}$

Therefore, the magnitude of force on $1^{\text{st}}$ particle is $64\text{N}$

The angle $\mathrm{\alpha }$ made by the force ${\overrightarrow{\mathrm{F}}}_1$ with horizontal line is,

  $\mathrm{\alpha }={\tan }^{-1}\left(\frac{\left|{\mathrm{F}}_{1\mathrm{y}}\right|}{\left|{\mathrm{F}}_{1\mathrm{x}}\right|}\right)$

  $\begin{array}{l}\text{Substitute }48.4\text{N for }{\mathrm{F}}_{1\mathrm{x}}\text{ and }41.9\text{N for }{\mathrm{F}}_{1\mathrm{y}}\\ \mathrm{\alpha }={\tan }^{-1}\left(\frac{\left|41.9\text{N}\right|}{\left|48.4\text{N}\right|}\right)\\ ={40.9}^{\circ }\end{array}$

The ${\mathrm{F}}_{1\mathrm{x}}$ is negative and ${\mathrm{F}}_{1\mathrm{y}}$ is negative, that implies ${\overrightarrow{\mathrm{F}}}_1$ lies in $3^{\text{rd}}$ quadrant

Therefore, the direction of net force on $1^{\text{st}}$ particle is at an angle of ${40.9}^{\circ }$ below the negative $\text{x}$ -axis.

From figure, the force acting on the $2^{\text{nd}}$ particle due to $3^{\text{rd}}$ particle is,

  ${\overrightarrow{\mathrm{F}}}_{23}={\mathrm{F}}_{23}\cos {60}^{\circ }(\text{<mover accent="true"><mn>i</mn><mo>^</mo></mover>})+{\mathrm{F}}_{23}\sin {60}^{\circ }(-\widehat{\mathrm{j}})$

Here, ${\mathrm{F}}_{23}$ is the force on $2^{\text{nd}}$ particle due to $3^{\text{rd}}$ particle.

From figure, the force acting on the $2^{\text{nd}}$ particle due to $1^{\text{st}}$ particle is,

  ${\overrightarrow{\mathrm{F}}}_{21}={\mathrm{F}}_{21}\cos {60}^{\circ }(\text{<mover accent="true"><mn>i</mn><mo>^</mo></mover>})$

Here, ${\mathrm{F}}_{12}$ is the force on $2^{\text{nd}}$ particle due to $1^{\text{st}}$ particle.

The net force acting on the $2^{\text{nd}}$ particle is,

  ${\overrightarrow{\mathrm{F}}}_2={\overrightarrow{\mathrm{F}}}_{21}+{\overrightarrow{\mathrm{F}}}_{23}$

  $=\left({\mathrm{F}}_{21}\cos {60}^{\circ }(\text{<mover accent="true"><mn>i</mn><mo>^</mo></mover>})\right)+\left({\mathrm{F}}_{23}\cos {60}^{\circ }(\text{<mover accent="true"><mn>i</mn><mo>^</mo></mover>})+{\mathrm{F}}_{23}\sin {60}^{\circ }(-\widehat{\mathrm{j}})\right)$

  $\begin{array}{l}\text{Substitute }48.4\text{N for }{\mathrm{F}}_{21}\text{ and }{\mathrm{F}}_{23}\\ {\overrightarrow{\mathrm{F}}}_2=\left((48.4\text{N})\cos {60}^{\circ }(\text{<mover accent="true"><mn>i</mn><mo>^</mo></mover>})\right)+\left((48.4\text{N})\cos {60}^{\circ }(\text{<mover accent="true"><mn>i</mn><mo>^</mo></mover>})+(48.4\text{N})\sin {60}^{\circ }(-\text{<mover accent="true"><mn>j</mn><mo>^</mo></mover>})\right)\\ {\overrightarrow{\mathrm{F}}}_2=(48.4\text{N})(\text{<mover accent="true"><mn>i</mn><mo>^</mo></mover>})+(41.9\text{N})(-\widehat{\mathrm{j}})\end{array}$

The magnitude of force acting on the $2^{\text{nd}}$ charged particle is,

  ${\mathrm{F}}_2=\sqrt{{\mathrm{F}}_{2\mathrm{x}}^2+{\mathrm{F}}_{2\mathrm{y}}^2}$

Here, ${\mathrm{F}}_{2\mathrm{x}}$ and ${\mathrm{F}}_{2\mathrm{y}}$ are the $\text{X}$ and $\text{Y}$ components of force ${\overrightarrow{\mathrm{F}}}_2$

Substitute $48.4\text{N}$ for ${\mathrm{F}}_{2\mathrm{x}}$ and $41.9\text{N}$ for ${\mathrm{F}}_{2\mathrm{y}}$

  $\begin{array}{l}{\mathrm{F}}_2=\sqrt{{(48.4\text{N})}^2+{(41.9\text{N})}^2}\hfill \\ {\mathrm{F}}_2=64\text{N}\hfill \end{array}$

Therefore, the magnitude of force on $2^{\text{nd}}$ particle is $64\text{N}$

The angle $\mathrm{\alpha }$ made by the force ${\overrightarrow{\mathrm{F}}}_2$ with horizontal line is,

  $\mathrm{\alpha }={\tan }^{-1}\left(\frac{\left|{\mathrm{F}}_{2\mathrm{y}}\right|}{\left|{\mathrm{F}}_{2\mathrm{x}}\right|}\right)$

Substitute $48.4\text{N}$ for ${\mathrm{F}}_{2\mathrm{x}}$ and $41.9\text{N}$ for ${\mathrm{F}}_{2\mathrm{y}}$

  $\begin{array}{cc}\mathrm{\alpha }& ={\tan }^{-1}\left(\frac{\left|41.9\text{N}\right|}{\left|48.4\text{N}\right|}\right)\\ & ={40.9}^{\circ }\end{array}$

The ${\mathrm{F}}_{2\mathrm{x}}$ is positive and ${\mathrm{F}}_{2\mathrm{y}}$ is negative, that implies ${\overrightarrow{\mathrm{F}}}_2$ lies in $4^{\text{th}}$ quadrant

Therefore, the direction of net force on $2^{\text{nd}}$ particle is at an angle of ${40.9}^{\circ }$ below the positive $\text{x}$ -axis