Chapter 24, Problem 29SP

Three point charges are placed at the following locations on the x-axis: $+2.0\mu \text{C at }x=0,-3.0\mu \text{C}$ at $x=40\text{ cm, }-5.0\mu \text{C at }x=120\text{ cm}$. Find the force (a) on the $-3.0\mu \text{C}$ charge, (b) on the $-5.0\mu \text{C}$ charge.

To determine

The electrostatic force on charge $-3.0\mu \text{C}$ if three point charges are placed at the following positions on the x-axis: $+2.0\mu \text{C}$ at $x=0$, $-3.0\mu \text{C}$ at $x=40\text{ cm}$, and $-5.0\mu \text{C}$ at $x=120\text{ cm}$.

Answer to Problem 29SP

Solution:

$-0.55\text{N}$

Explanation of Solution

Given data:

Charge on point charge at $x=0$ is $+2.0\mu \text{C}$.

Charge on point charge at $x=40\text{ cm}$ is $-3.0\mu \text{C}$.

Charge on point charge at $x=120\text{ cm}$ is $-5.0\mu \text{C}$.

Separation between $+2.0\mu \text{C}$ and $-3.0\mu \text{C}$ is $40\text{cm}$.

Separation between $-3.0\mu \text{C}$ and $-5.0\mu \text{C}$ is $80\text{cm}$.

Formula used:

Write the expression for the electrostatic force:

$F=\frac{\left|q_1\right|\left|q_2\right|}{4\pi {\epsilon }_o{r}^2}$

Here, $F$ is the electrostatic force, $q_1$ charge on first point charge, $q_2$ charge on other point charge, ${\epsilon }_0$ permittivity of free space, and $r$ is the separation between the point charges.

Explanation:

Draw the diagram of the schematic arrangement:

Here, $F_{E2}$ represents the force acting on charge $-3.0\mu \text{C}$ due to charge $+2.0\mu \text{C}$ and $F_{E5}$ represents the force acting on charge $-3.0\mu \text{C}$ due to charge $+2.0\mu \text{C}$.

Consider the expression for the electrostatic force on $-3.0\mu \text{C}$ by $+2.0\mu \text{C}$:

$F_{E2}=\frac{1}{4\pi {\epsilon }_0}\frac{\left|q_1\right|\left|q_2\right|}{r^2}$

Substitute $+2.0\mu \text{C}$ for $q_1$, $-3.0\mu \text{C}$ for $q_2$, $40\text{cm}$ for $r$, and $9\times {10}^9\text{N}{\text{m}}^2{\text{/C}}^2$ for $\frac{1}{4\pi {\epsilon }_0}$

$\begin{array}{c}{F}_1=\left(9\times {10}^9\text{N}\cdot {\text{m}}^2{\text{/C}}^2\right)\frac{\left|\left(+2.0\mu \text{C}\right)\right|\left|\left(-3.0\mu \text{C}\right)\right|}{{\left(40\text{ cm}\right)}^2}\\ =\left(9\times {10}^9\text{N}\cdot {\text{m}}^2{\text{/C}}^2\right)\frac{\left(2.0\mu \text{C}\right)\left(\frac{1\text{ C}}{{10}^6\mu \text{C}}\right)\left(-3.0\mu \text{C}\right)\left(\frac{1\text{ C}}{{10}^6\mu \text{C}}\right)}{{\left(40\text{ cm}\left(\frac{1\text{ m}}{{10}^2\text{ cm}}\right)\right)}^2}\\ =-0.34\text{N}\end{array}$

Similarly, consider the expression for the electrostatic force on $-3.0\mu \text{C}$ by $-5.0\mu \text{C}$:

$F_{E5}=\frac{1}{4\pi {\epsilon }_0}\frac{\left|q_2\right|\left|q_3\right|}{r^2}$

Substitute $-3.0\mu \text{C}$ for $q_2$, $-5.0\mu \text{C}$ for $q_3$, $40\text{cm}$ for $r$, and $9\times {10}^9\text{N}{\text{m}}^2{\text{/C}}^2$ for $\frac{1}{4\pi {\epsilon }_0}$

$\begin{array}{c}{F}_{E5}=\left(9\times {10}^9\text{N}\cdot {\text{m}}^2{\text{/C}}^2\right)\frac{\left|\left(-3.0\mu \text{C}\right)\right|\left|\left(-5.0\mu \text{C}\right)\right|}{{\left(80\text{ cm}\left(\frac{1\text{ m}}{{10}^2\text{cm}}\right)\right)}^2}\\ =\left(9\times {10}^9\text{N}\cdot {\text{m}}^2{\text{/C}}^2\right)\frac{\left(3.0\mu \text{C}\left(\frac{1\text{ C}}{{10}^6\mu \text{C}}\right)\right)\left(5.0\mu \text{C}\left(\frac{1\text{ C}}{{10}^6\mu \text{C}}\right)\right)}{{\left(80\text{ cm}\left(\frac{1\text{ m}}{{10}^2\text{cm}}\right)\right)}^2}\\ =0.21\text{ N}\end{array}$

Understand that the force acting along x-axis on the charge $-3.0\mu \text{C}$ will be the sum of the forces due to the charges $-5.0\mu \text{C}$ and $+2.0\mu \text{C}$ because forces are in same direction.

$F=F_{E2}+F_{E5}$

Substitute $0.34\text{ N}$ for $F_1$ and $0.21\text{ N}$ for $F_2$

$\begin{array}{c}F=\left(0.34\text{ N}\right)+\left(0.21\text{ N}\right)\\ =0.55\text{ N}\end{array}$

The electrostatic force on $-3.0\mu \text{C}$ charge is $0.55\text{ N}$.

Understand that the force acting along negative direction. Therefore, the force acting on the charge $-3.0\mu \text{C}$ is $-0.55\text{ N}$.

Conclusion:

The electrostatic forceon $-3.0\mu \text{C}$ charge is $-0.55\text{ N}$.

To determine

The electrostatic force on charge $-5.0\mu \text{C}$ if three point charges are placed at the following positions on the x-axis: $+2.0\mu \text{C}$ at $x=0$, $-3.0\mu \text{C}$ at $x=40\text{ cm}$, and $-5\mu \text{C}$ at $x=120\text{ cm}$.

Answer to Problem 29SP

Solution:

$0.15\text{N}$

Explanation of Solution

Given data:

Charge on point charge at $x=0$ is $+2.0\mu \text{C}$.

Charge on point charge at $x=40\text{ cm}$ is $-3.0\mu \text{C}$.

Charge on point charge at $x=120\text{ cm}$ is $-5.0\mu \text{C}$.

Separation between $+2.0\mu \text{C}$ and $-3.0\mu \text{C}$ is $40\text{ cm}$.

Separation between $-3.0\mu \text{C}$ and $-5.0\mu \text{C}$ is $80\text{ cm}$.

Formula used:

Write the expression for the electrostatic force:

$F=\frac{\left|q_1\right|\left|q_2\right|}{4\pi {\epsilon }_o{r}^2}$

Here, $F$ is the electrostatic force, $q_1$ charge on first point charge, $q_2$ charge on other point charge, ${\epsilon }_0$ permittivity of free space, and $r$ is the separation between the point charges.

Explanation:

Draw the schematic arrangement of the charges as shown below:

Consider the expression for the electrostatic force on $-5.0\mu \text{C}$ by $+2.0\mu \text{C}$:

$F_{E2}=\frac{1}{4\pi {\epsilon }_0}\frac{\left|q_1\right|\left|q_3\right|}{{\left(r_{13}\right)}^2}$

The distance between the charges $-5.0\mu \text{C}$ and $+2.0\mu \text{C}$ is $r_{13}$.

And,

$r_{13}=r_{12}+r_{23}$

Substitute $40\text{ cm}$ for $r_{12}$ and $80\text{ cm}$ for $r_{13}$

$\begin{array}{c}{r}_{13}=40\text{ cm}+80\text{ cm}\\ \text{=120 cm}\end{array}$

Substitute, $+2.0\mu \text{C}$ for $q_1$, $-5.0\mu \text{C}$ for $q_3$, $120\text{cm}$ for $r_{13}$, and $9\times {10}^9\text{N}{\text{m}}^2{\text{/C}}^2$ for $\frac{1}{4\pi {\epsilon }_0}$

$\begin{array}{c}{F}_{E2}=\left(9\times {10}^9\text{ N}\cdot {\text{m}}^2{\text{/C}}^2\right)\frac{\left|+2.0\mu \text{C}\right|\left|-5.0\mu \text{C}\right|}{{\left(120\text{ cm}\right)}^2}\\ =\left(9\times {10}^9\text{ N}\cdot {\text{m}}^2{\text{/C}}^2\right)\frac{\left(2.0\mu \text{C}\right)\left(\frac{1\text{ C}}{{10}^6\mu \text{C}}\right)\left(-5.0\mu \text{C}\right)\left(\frac{1\text{ C}}{{10}^6\mu \text{C}}\right)}{{\left(120\text{ cm}\left(\frac{1\text{}\text{ m}}{{10}^2\text{ cm}}\right)\right)}^2}\\ =0.063\text{N}\end{array}$

Similarly, consider the expression for the electrostatic force on $-5.0\mu \text{C}$ by $-3.0\mu \text{C}$:

$F_{E3}=\frac{1}{4\pi {\epsilon }_0}\frac{\left|q_2\right|\left|q_3\right|}{r_{23}{}^2}$

Substitute $-5.0\mu \text{C}$ for $q_3$, $-3.0\mu \text{C}$ for $q_2$, $80\text{ cm}$ for $r$, and $9\times {10}^9\text{ N}\cdot {\text{m}}^2{\text{/C}}^2$ for $\frac{1}{4\pi {\epsilon }_0}$

$\begin{array}{c}{F}_{E3}=\left(9\times {10}^9\text{ N}\cdot {\text{m}}^2{\text{/C}}^2\right)\frac{\left|-5.0\mu \text{C}\right|\left|-3.0\mu \text{C}\right|}{{\left(80\text{ cm}\right)}^2}\\ =\left(9\times {10}^9\text{ N}\cdot {\text{m}}^2{\text{/C}}^2\right)\frac{\left(5.0\mu \text{C}\right)\left(\frac{1\text{ C}}{{10}^6\mu \text{C}}\right)\left(3.0\mu \text{C}\right)\left(\frac{1\text{ C}}{{10}^6\mu \text{C}}\right)}{{\left(80\text{ cm}\left(\frac{1\text{ m}}{{10}^2\text{ cm}}\right)\right)}^2}\\ =0.21\text{ N}\end{array}$

From the diagram provided above, the net electrostatic force on charge $-5.0\mu \text{C}$ is

$F=F_{E3}-F_{E2}$

Substitute $-0.063\text{N}$ for $F_{E3}$ and $0.21\text{N}$ for $F_{E2}$

$\begin{array}{c}F=0.21\text{ N}-0.063\text{ N}\\ =0.15\text{ N}\end{array}$

Conclusion:

The electrostatic force on $-5.0\mu \text{C}$ charge is $0.15\text{N}$.