Chapter 16, Problem 17P

To determine

To find: the direction and magnitude on each due to the other two.

Explanation of Solution

Given:

The given figure is shown below.

  

Formula used:

From coulomb's law, electrostatic force for two charges is given as,

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

Here, $\mathrm{F}$ is the electrostatic force, ${\mathrm{Q}}_1$ and ${\mathrm{Q}}_2$ are two charges, which are separated by distance $\mathrm{r}$ and $\mathrm{k}$ is the proportionality constant.

The net force of electrostatic is given as,

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

Here, ${\mathrm{F}}_{\text{x}}$ is the net force in $\text{x}$ -direction and ${\mathrm{F}}_{\text{y}}$ is the net force in $\text{y}$ -direction.

Calculation:

The direction of the force exerted on the three charges are shown as,

  

From diagram at point A the electrostatic force of charge ${\mathrm{Q}}_1$ and ${\mathrm{Q}}_2$ is given as,

  ${\mathrm{F}}_{12}=\mathrm{k}\frac{{\mathrm{Q}}_1{\mathrm{Q}}_2}{{\mathrm{r}}^2}$

Substitute the value, $4.0\times {10}^{-6}\text{C}\text{for}{\mathrm{Q}}_1,-8.0\times {10}^{-6}\text{Cfor}{\mathrm{Q}}_2,9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $\mathrm{k}$ and $1.20\text{m}$ for $\mathrm{r}$ in the above expression,

  $\begin{array}{l}{\mathrm{F}}_{12}=\mathrm{k}\frac{{\mathrm{Q}}_1{\mathrm{Q}}_2}{{\mathrm{r}}^2}\\ =\left(9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\frac{\left(4.0\times {10}^{-6}\text{C}\right)\left(-8.0\times {10}^{-6}\text{C}\right)}{{(1.20\text{m})}^2}\\ =-2\times {10}^{-2}\text{N}\end{array}$

From diagram at point $\mathrm{A}$ the electrostatic force of charge ${\mathrm{Q}}_1$ and ${\mathrm{Q}}_3$ is given as,

  ${\mathrm{F}}_{13}=\mathrm{k}\frac{{\mathrm{Q}}_1{\mathrm{Q}}_3}{{\mathrm{r}}^2}$

  $4.0\times {10}^{-6}\text{C}\text{for}{\mathrm{Q}}_1,-6.0\times {10}^{-6}\text{Cfor}{\mathrm{Q}}_3,9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $\mathrm{k}$ and $1.20\text{m}$ for $\mathrm{r}$ in the above expression,

  $\begin{array}{cc}{\mathrm{F}}_{13}& =\mathrm{k}\frac{{\mathrm{Q}}_1{\mathrm{Q}}_3}{{\mathrm{r}}^2}\\ & =\left(9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\frac{\left(4.0\times {10}^{-6}\text{C}\right)\left(-6.0\times {10}^{-6}\text{C}\right)}{{(1.20\text{m})}^2}\\ & =-1.50\times {10}^{-2}\text{N}\end{array}$

From diagram at point $\mathrm{A}$ net force in x-direction is given as,

  ${\mathrm{F}}_{\text{x}}={\mathrm{F}}_{13}\cos {60}^{\circ }-{\mathrm{F}}_{12}\cos {60}^{\circ }$

Substitute the value of ${\mathrm{F}}_{12}$ and ${\mathrm{F}}_{13}$ in the above expression,

  $\begin{array}{cc}{\mathrm{F}}_{\mathrm{x}}& ={\mathrm{F}}_{13}\cos {60}^{\circ }-{\mathrm{F}}_{12}\cos {60}^{\circ }\\ & =\left(-1.50\times {10}^{-2}\text{N}\right)\cos {60}^{\circ }-\left(-2\times {10}^{-2}\text{N}\right)\cos {60}^{\circ }\\ & =-7.50\times {10}^{-3}\text{N}+10\times {10}^{-3}\text{N}\\ & =25.0\times {10}^{-4}\text{N}\end{array}$

From diagram at point $\mathrm{A}$ net force in y-direction is given as,

  $\begin{array}{cc}{\mathrm{F}}_{\mathrm{y}}& =-{\mathrm{F}}_{13}\sin {60}^{\circ }-{\mathrm{F}}_{12}\sin {60}^{\circ }\\ & =-\left(-1.50\times {10}^{-2}\text{N}\right)\sin {60}^{\circ }-\left(-2\times {10}^{-2}\text{N}\right)\sin {60}^{\circ }\\ & =1.30\times {10}^{-2}\text{N}+1.732\times {10}^{-2}\text{N}\\ & =303.2\times {10}^{-4}\text{N}\end{array}$

The net force of electrostatic at point A is given as,

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

Substitute the value of ${\mathrm{F}}_{\text{x}}$ and ${\mathrm{F}}_{\text{y}}$ in the above expression,

  $\begin{array}{cc}{\mathrm{F}}_{\text{net}}& =\sqrt{{\mathrm{F}}_{\text{x}}^2+{\mathrm{F}}_{\text{y}}^2}\\ & =\sqrt{{\left(25.0\times {10}^{-4}\text{N}\right)}^2+{\left(303.2\times {10}^{-4}\text{N}\right)}^2}\\ & =\sqrt{6.25\times {10}^{-6}{\text{N}}^2+919.30\times {10}^{-6}{\text{N}}^2}\\ & =\sqrt{925.55\times {10}^{-6}{\text{N}}^2}\\ & =3.04\times {10}^{-2}\text{N}\end{array}$

The direction of net forces on ${\mathrm{Q}}_1$ due to ${\mathrm{Q}}_2$ and ${\mathrm{Q}}_3$ is given as,

  $\tan \mathrm{\theta }=\frac{{\mathrm{F}}_{\text{y}}}{{\mathrm{F}}_{\text{x}}}$

Substitute the value of ${\mathrm{F}}_{\text{x}}$ and ${\mathrm{F}}_{\text{y}}$ in the above expression,

  $\begin{array}{cc}\tan \mathrm{\theta }& =\frac{{\mathrm{F}}_{\mathrm{y}}}{{\mathrm{F}}_{\mathrm{x}}}\\ & =\frac{\left(303.2\times {10}^{-4}\text{N}\right)}{\left(25.0\times {10}^{-4}\text{N}\right)}\\ & =12.128\\ \mathrm{\theta }& ={\tan }^{-1}(12.128)\\ & ={85.28}^{\circ }\end{array}$

Hence, the direction of net forces on ${\mathrm{Q}}_1$ due to ${\mathrm{Q}}_2$ and ${\mathrm{Q}}_3$ is ${85.28}^{\circ }$

From diagram at point $\mathrm{B}$ the electrostatic force of charge ${\mathrm{Q}}_1$ and ${\mathrm{Q}}_2$ is given as,

  ${\mathrm{F}}_{21}=\mathrm{k}\frac{{\mathrm{Q}}_1{\mathrm{Q}}_2}{{\mathrm{r}}^2}$

Substitute the value, $4.0\times {10}^{-6}\text{Cfor}{\mathrm{Q}}_1,-8.0\times {10}^{-6}\text{Cfor}{\mathrm{Q}}_2,9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $\mathrm{k}$ and $1.20\text{m}$ for $\mathrm{r}$ in the above expression,

  $\begin{array}{cc}{\mathrm{F}}_{21}& =\mathrm{k}\frac{{\mathrm{Q}}_1{\mathrm{Q}}_2}{{\mathrm{r}}^2}\\ & =\left(9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\frac{\left(4.0\times {10}^{-6}\text{C}\right)\left(-8.0\times {10}^{-6}\text{C}\right)}{{(1.20\text{m})}^2}\\ & =-2\times {10}^{-2}\text{N}\end{array}$

From diagram at point $\mathrm{B}$ the electrostatic force of charge ${\mathrm{Q}}_3$ and ${\mathrm{Q}}_2$ is given as,

  ${\mathrm{F}}_{23}=\mathrm{k}\frac{{\mathrm{Q}}_2{\mathrm{Q}}_3}{{\mathrm{r}}^2}$

Substitute the value, $-6.0\times {10}^{-6}\text{C}\text{for}{\mathrm{Q}}_3,-8.0\times {10}^{-6}\text{Cfor}{\mathrm{Q}}_2,9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $\mathrm{k}$ and $1.20\text{m}$ for $\mathrm{r}$ in the above expression,

  $\begin{array}{cc}{\mathrm{F}}_{23}& =\mathrm{k}\frac{{\mathrm{Q}}_2{\mathrm{Q}}_3}{{\mathrm{r}}^2}\\ & =\left(9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\frac{\left(-8.0\times {10}^{-6}\text{C}\right)\left(-6.0\times {10}^{-6}\text{C}\right)}{{(1.20\text{m})}^2}\\ & =0.30\text{N}\end{array}$

From diagram at point $\mathrm{B}$ net force in $\mathrm{x}$ -direction is given as,

  ${\mathrm{F}}_{\text{x}}={\mathrm{F}}_{21}\cos {60}^{\circ }+{\mathrm{F}}_{23}$

Substitute the value of ${\mathrm{F}}_{21}$ and ${\mathrm{F}}_{23}$ in the above expression,

  $\begin{array}{cc}{\mathrm{F}}_{\text{x}}& ={\mathrm{F}}_{21}\cos {60}^{\circ }+{\mathrm{F}}_{23}\\ & =\left(-2\times {10}^{-2}\text{N}\cos {60}^{\circ }\right)+(0.30\text{N})\\ & =0.29\text{N}\end{array}$

From diagram at point $\mathrm{B}$ net force in $\mathrm{y}$ -direction is given as,

  ${\mathrm{F}}_{\text{y}}={\mathrm{F}}_{21}\sin {60}^{\circ }$

Substitute the value of ${\mathrm{F}}_{21}$ in the above expression,

  $\begin{array}{cc}{\mathrm{F}}_{\mathrm{y}}& ={\mathrm{F}}_{21}\sin {60}^{\circ }\\ & =\left(-2\times {10}^{-2}\text{N}\right)\sin {60}^{\circ }\\ & =-0.0173\text{N}\end{array}$

The net force of electrostatic at point $\mathrm{B}$ is given as,

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

Substitute the value of ${\mathrm{F}}_{\text{x}}$ and ${\mathrm{F}}_{\text{y}}$ in the above expression,

  $\begin{array}{cc}{\mathrm{F}}_{\text{net}}& =\sqrt{{\mathrm{F}}_{\text{x}}^2+{\mathrm{F}}_{\text{y}}^2}\\ & =\sqrt{{(0.29\text{N})}^2+{(-0.0173\text{N})}^2}\\ & =\sqrt{0.0844{\text{N}}^2}\\ & =0.29\text{N}\end{array}$

Hence, magnitude of net force at point $\text{B}$ is $0.29\text{N}$

The direction of net forces on ${\mathrm{Q}}_2$ due to ${\mathrm{Q}}_1$ and ${\mathrm{Q}}_3$ is given as,

  $\tan \mathrm{\theta }=\frac{{\mathrm{F}}_{\text{y}}}{{\mathrm{F}}_{\text{x}}}$

Substitute the value of ${\mathrm{F}}_{\mathrm{x}}$ and ${\mathrm{F}}_{\mathrm{y}}$ in the above expression,

  $\begin{array}{cc}\tan \mathrm{\theta }& =\frac{{\mathrm{F}}_{\mathrm{y}}}{{\mathrm{F}}_{\text{x}}}\\ & =\frac{(-0.0137\text{N})}{(0.29\text{N})}\\ & =-0.0596\\ \mathrm{\theta }& ={\tan }^{-1}(-0.0596)\\ & =-{3.41}^{\circ }\end{array}$

Here, the negative sign shows that the direction of the net force at point $\mathrm{B}$ is in third quadrant from x -axis.

Hence, the direction of net forces on ${\mathrm{Q}}_2$ due to ${\mathrm{Q}}_1$ and ${\mathrm{Q}}_3$ is ${3.41}^{\circ }$

From diagram at point $\mathrm{C}$ the electrostatic force of charge ${\mathrm{Q}}_3$ and ${\mathrm{Q}}_1$ is given as,

  ${\mathrm{F}}_{31}=\mathrm{k}\frac{{\mathrm{Q}}_3{\mathrm{Q}}_1}{{\mathrm{r}}^2}$

Substitute the value, $4.0\times {10}^{-6}{\text{C}}^{\text{for}}{\mathrm{Q}}_1,-6.0\times {10}^{-6}{\text{C}}^{\text{for}}{\mathrm{Q}}_3,9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $\mathrm{k}$ and $1.20\text{m}$ for $\mathrm{r}$ in the above expression,

  $\begin{array}{cc}{\mathrm{F}}_{31}& =\mathrm{k}\frac{{\mathrm{Q}}_3{\mathrm{Q}}_1}{{\mathrm{r}}^2}\\ & =\left(9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\frac{\left(-6.0\times {10}^{-6}\text{C}\right)\left(4.0\times {10}^{-6}\text{C}\right)}{{(1.20\text{m})}^2}\\ & =-1.50\times {10}^{-2}\text{N}\end{array}$

From diagram at point $\mathrm{C},$ the electrostatic force of charge ${\mathrm{Q}}_3$ and ${\mathrm{Q}}_2$ is given as,

  ${\mathrm{F}}_{32}=\mathrm{k}\frac{{\mathrm{Q}}_3{\mathrm{Q}}_2}{{\mathrm{r}}^2}$

Substitute the value, $-6.0\times {10}^{-6}{\text{C}}^{\text{for}}{\mathrm{Q}}_3,-8.0\times {10}^{-6}{\text{C}}^{\text{for}}{\mathrm{Q}}_2,9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $\mathrm{k}$ and $1.20\text{m}$ for $\mathrm{r}$ in the above expression,

  $\begin{array}{cc}{\mathrm{F}}_{32}& =\mathrm{k}\frac{{\mathrm{Q}}_3{\mathrm{Q}}_2}{{\mathrm{r}}^2}\\ & =\left(9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\frac{\left(-6.0\times {10}^{-6}\text{C}\right)\left(-8.0\times {10}^{-6}\text{C}\right)}{{(1.20\text{m})}^2}\\ & =0.30\text{N}\end{array}$

From diagram at point $\mathrm{C},$ net force in $\mathrm{x}$ -direction is given as,

  ${\mathrm{F}}_{\text{x}}=-{\mathrm{F}}_{31}\cos {60}^{\circ }-{\mathrm{F}}_{32}$

Substitute the value of $\text{C},$ and ${\mathrm{F}}_{32}$ in the above expression,

  ${\mathrm{F}}_{\text{y}}={\mathrm{F}}_{31}\sin {60}^{\circ }$

From diagram at point $\text{C},$ net force in $\text{y}$ -direction is given as,

  ${\mathrm{F}}_{\text{y}}={\mathrm{F}}_{31}\sin {60}^{\circ }$

Substitute the value of ${\mathrm{F}}_{31}$ in the above expression,

  $\begin{array}{cc}{\mathrm{F}}_{\mathrm{y}}& ={\mathrm{F}}_{31}\sin {60}^{\circ }\\ & =\left(-1.50\times {10}^{-2}\text{N}\right)\sin {60}^{\circ }\\ & =-0.013\text{N}\end{array}$

The net force of electrostatic at point $\mathrm{C}$ is given as,

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

Substitute the value of ${\mathrm{F}}_{\text{x}}$ and ${\mathrm{F}}_{\text{y}}$ in the above expression,

  $\begin{array}{cc}{\mathrm{F}}_{\text{net}}& =\sqrt{{\mathrm{F}}_{\text{x}}^2+{\mathrm{F}}_{\text{y}}^2}\\ & =\sqrt{{(-0.2925\text{N})}^2+{(-0.013\text{N})}^2}\\ & =\sqrt{0.08572{\text{N}}^2}\\ & =0.291\text{N}\end{array}$

Hence, magnitude of net force at point $\mathrm{C}$ is $0.291\text{N}$

The direction of net forces on ${\mathrm{Q}}_3$ due to ${\mathrm{Q}}_1$ and ${\mathrm{Q}}_2$ is given as,

  $\tan \mathrm{\theta }=\frac{{\mathrm{F}}_{\mathrm{y}}}{{\mathrm{F}}_{\mathrm{x}}}$

Substitute the value of ${\mathrm{F}}_{\text{x}}$ and ${\mathrm{F}}_{\text{y}}$ in the above expression,

  $\begin{array}{cc}\tan \mathrm{\theta }& =\frac{{\mathrm{F}}_{\mathrm{y}}}{{\mathrm{F}}_{\mathrm{x}}}\\ & =\frac{(-0.013\text{N})}{(-0.2925\text{N})}\\ & =0.0444\\ \mathrm{\theta }& ={\tan }^{-1}(0.0444)\\ & ={2.54}^{\circ }\end{array}$

Hence, the direction of net forces on ${\mathrm{Q}}_3$ due to ${\mathrm{Q}}_1$ and ${\mathrm{Q}}_2$ is ${2.54}^{\circ }$