Chapter 23, Problem 38PQ

Given the arrangement of charged particles in Figure P23.38, find the net electrostatic force on the 5.65-μC charged particle.

To determine

The net-electrostatic force on $5.65\mu \text{C}$ charged particle.

Answer to Problem 38PQ

The net-electrostatic force on $5.65\mu \text{C}$ charged particle is $\left(876\widehat{i}-853\widehat{j}\right)\text{N}$.

Explanation of Solution

The diagram for the charges.

Write the expression for Coulomb’s law.

${\overrightarrow{F}}_E=\frac{k{q}_1{q}_2}{r^2}\widehat{r}$ (I)

Here, $F_E$ is the electrostatic force between the particles, $k$ is the Coulomb’s constant, $q_1$ is the charge of particle 1, $q_2$ is the charge of particle 2, $r$ is the distance between the centers of the particles.

Write the distance between the charges $q_1$ and $q_2$ from above figure.

  $\begin{array}{c}{r}_{q_2}=\sqrt{{\left(-2.00\text{cm}\frac{{10}^{-2}\text{m}}{1.0\text{cm}}-\text{1}\text{.00}\text{cm}\frac{{10}^{-2}\text{m}}{1.0\text{cm}}\right)}^2+{\left(1.00\text{cm}\frac{{10}^{-2}\text{m}}{1.0\text{cm}}\right)}^2}\\ =\sqrt{{\left(-0.0200\text{m}-0.0\text{100}\text{m}\right)}^2+{\left(0.0100\text{m}\right)}^2}\\ =\sqrt{1.00\times {10}^{-3}{\text{m}}^{\text{2}}}\\ r_{q_2}^2=1.00\times {10}^{-3}{\text{m}}^{\text{2}}\end{array}$

Here, $r_{q_2}$ is the distance between charges $q_1$ and $q_2$.

Write the equation for distance between the charges $q_1$ and $q_3$ from above figure.

  $\begin{array}{c}{r}_{q_3}=\sqrt{{\left(-2.00\text{cm}\frac{{10}^{-2}\text{m}}{1.0\text{cm}}\right)}^2+{\left(1.00\text{cm}\frac{{10}^{-2}\text{m}}{1.0\text{cm}}\right)}^2}\\ =\sqrt{{\left(-0.0200\text{m}\right)}^2+{\left(0.0100\text{m}\right)}^2}\\ =\sqrt{5.00\times {10}^{-4}{\text{m}}^{\text{2}}}\\ r_{q_3}^2=5.00\times {10}^{-4}{\text{m}}^{\text{2}}\end{array}$

Here, $r_{q_3}$ is the distance between charges $q_1$ and $q_3$.

Write the equation for angle between lines of particle with charge $q_2$, $q_1$ and x-axis form the figure.

  $\begin{array}{c}{\theta }_{q_2}={\tan }^{-1}\left(\left|\frac{\text{1}\text{.00}\text{cm}\frac{{10}^{-2}\text{m}}{1.0\text{cm}}}{\text{3}\text{.00}\text{cm}\frac{{10}^{-2}\text{m}}{1.0\text{cm}}}\right|\right)\\ ={\tan }^{-1}\left(\left|\frac{0.0\text{100}\text{m}}{0.03\text{00}\text{m}}\right|\right)\\ =18.435°\end{array}$

Here, ${\theta }_{q_2}$ is the angle between lines of particle with charge $q_2$, $q_1$ and x-axis.

Write the equation for angle between lines of particle with charge $q_3$, $q_1$ and x-axis form the figure.

  $\begin{array}{c}{\theta }_{q_3}={\tan }^{-1}\left(\left|\frac{\text{1}\text{.00}\text{cm}\frac{{10}^{-2}\text{m}}{1.0\text{cm}}}{\text{2}\text{.00}\text{cm}\frac{{10}^{-2}\text{m}}{1.0\text{cm}}}\right|\right)\\ ={\tan }^{-1}\left(\left|\frac{0.0\text{100}\text{m}}{0.02\text{00}\text{m}}\right|\right)\\ =26.565°\end{array}$

Here, ${\theta }_{q_3}$ is the angle between lines of particle with charge $q_3$, $q_1$ and x-axis.

Since, both the particles with charges $q_2$ and $q_1$ are positive, the force between will be repulsive and it will result a negative x-component and a negative y-component.

Write the expression to find the electrostatic force by particles $q_2$ on $q_1$.

  ${\overrightarrow{F}}_{q_2}=\frac{k{q}_1{q}_2}{r_{q_2}{}^2}\left(-\cos {\theta }_{{}_{q_2}}\widehat{i}-\sin {\theta }_{{}_{q_2}}\widehat{j}\right)$

Here, ${\overrightarrow{F}}_{q_2}$ is the electrostatic force by particles $q_2$ on $q_1$.

Substitute $8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$, $5.65\mu \text{C}$ for $q_1$ and $10.33\mu \text{C}$ for $q_2$, $1.00\times {10}^{-3}{\text{m}}^{\text{2}}$ for $r_{q_2}^2$ and $18.435°$ for ${\theta }_{q_2}$ to find the electrostatic force by particles $q_2$ on $q_1$.

  $\begin{array}{c}{\overrightarrow{F}}_{q_2}=\frac{\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left|\left(5.65\mu \text{C}\frac{{10}^{-6}\text{C}}{1.0\mu \text{C}}\right)\left(10.33\mu \text{C}\frac{{10}^{-6}\text{C}}{1.0\mu \text{C}}\right)\right|}{1.00\times {10}^{-3}{\text{m}}^{\text{2}}}\\ \left(-\cos \left(18.435°\right)\widehat{i}-\sin \left(18.435°\right)\widehat{j}\right)\\ =\frac{\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(5.65\times {10}^{-6}\text{C}\right)\left(10.33\times {10}^{-6}\text{C}\right)}{1.00\times {10}^{-3}{\text{m}}^{\text{2}}}\\ \left(-\cos \left(18.435°\right)\widehat{i}-\sin \left(18.435°\right)\widehat{j}\right)\\ =\left(-498\widehat{i}-166\widehat{j}\right)\text{N}\end{array}$

Since, the particle $q_3$ is negative and $q_1$ is positive, the force between will be attractive and it will result a positive x-component and a negative y-component.

Write the expression to find the electrostatic force by particles $q_3$ on $q_1$.

  ${\overrightarrow{F}}_{q_3}=\frac{k{q}_1{q}_3}{r_{q_3}{}^2}\left(\cos {\theta }_{{}_{q_3}}\widehat{i}-\sin {\theta }_{{}_{q_3}}\widehat{j}\right)$

Here, ${\overrightarrow{F}}_{q_3}$ is the electrostatic force by particles $q_3$ on $q_1$.

Substitute $8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$, $5.65\mu \text{C}$ for $q_1$ and $-15.12\mu \text{C}$ for $q_3$, $5.00\times {10}^{-4}{\text{m}}^{\text{2}}$ for $r_{q_3}^2$ and $26.565°$ for ${\theta }_{q_3}$ to find the electrostatic force by particles $q_3$ on $q_1$.

  $\begin{array}{c}{\overrightarrow{F}}_{q_3}=\frac{\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left|\left(5.65\mu \text{C}\frac{{10}^{-6}\text{C}}{1.0\mu \text{C}}\right)\left(-15.12\mu \text{C}\frac{{10}^{-6}\text{C}}{1.0\mu \text{C}}\right)\right|}{5.00\times {10}^{-4}{\text{m}}^{\text{2}}}\\ \left(\cos \left(26.565°\right)\widehat{i}-\sin \left(26.565°\right)\widehat{j}\right)\\ =\frac{\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(5.65\times {10}^{-6}\text{C}\right)\left(15.12\times {10}^{-6}\text{C}\right)}{5.00\times {10}^{-4}{\text{m}}^{\text{2}}}\\ \left(\cos \left(26.565°\right)\widehat{i}-\sin \left(26.565°\right)\widehat{j}\right)\\ =\left(1.374\times {10}^3\widehat{i}-687\widehat{j}\right)\text{N}\end{array}$

The net-electrostatic force on $5.65\mu \text{C}$ charged particle will be the sum of the forces ${\overrightarrow{F}}_{q_2}$ and ${\overrightarrow{F}}_{q_3}$.

Write the expression to find the net-electrostatic force on $5.65\mu \text{C}$ charged particle.

  $\overrightarrow{F}={\overrightarrow{F}}_{q_2}+{\overrightarrow{F}}_{q_3}$

Here, $\overrightarrow{F}$ is the net-electrostatic force on $5.65\mu \text{C}$ charged particle.

Conclusion:

Substitute $\left(-498\widehat{i}-166\widehat{j}\right)\text{N}$ for ${\overrightarrow{F}}_{q_2}$ and $\left(1.374\times {10}^3\widehat{i}-687\widehat{j}\right)\text{N}$ for ${\overrightarrow{F}}_{q_3}$ to find the net-electrostatic force on net-electrostatic force on $5.65\mu \text{C}$ charged particle.

  $\begin{array}{c}\overrightarrow{F}=\left(-498\widehat{i}-166\widehat{j}\right)\text{N}+\left(1.374\times {10}^3\widehat{i}-687\widehat{j}\right)\text{N}\\ =\left(876\widehat{i}-853\widehat{j}\right)\text{N}\end{array}$

Therefore, The net-electrostatic force on $5.65\mu \text{C}$ charged particle is $\left(876\widehat{i}-853\widehat{j}\right)\text{N}$.