Chapter 12, Problem 1SP

Three positive charges are located along a line, as in the diagram. The 0.14 C charge at points A is 2.5 m to the left of the 0.06 C charge at point B, and the 0.09 C charge at point C is 1.5 m to the right of point B.

a.    What is the magnitude of the force exerted on the 0.06 C charge by the 0.14 C charge?

b.    What is the magnitude of the force exerted on the 0.06 C charge by the 0.09 C charge?

c.    What is the net force exerted on the 0.06 C charge by the other two charges?

d.    If we regard the 0.06 C charge as a test charge used to probe the strength of the electric field produced by the other two charges, what are the magnitude and direction of the electric field at point B?

e.    If the 0.06 C charge at point B is replaced by a –0.17 C charge, what are the magnitude and direction of the electrostatic force exerted on this new charge? (Use the electric field value to find this force.)

To determine

The magnitude of the force on $0.06\text{C}$ charge by $0.14\text{C}$.

Answer to Problem 1SP

The magnitude of the force on $0.06\text{C}$ charge by $0.14\text{C}$ is $1.21\times {10}^7\text{N}$.

Explanation of Solution

Given Info: The distance between the charges at A and B is $2.5\text{m}$, the charge at point A is $+0.14\text{C}$, the charge at point B is $+0.06\text{C}$.

The diagram showing the charges

Write the expression to find the electrostatic force.

$F=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r^2}$

Here,

$F$ is the electrostatic force

$q_1$ is the first charge

$q_2$ is the second charge

$r$ is the distance between the charges

${\epsilon }_0$ is the permittivity of free space

Substitute $8.85\times {10}^{-12}\text{F/m}$ for ${\epsilon }_0$, $+0.06\text{C}$ for $q_1$, $+0.14\text{C}$ for $q_2$, $2.5\text{m}$ for $r$ to find the electrostatic force.

$\begin{array}{c}F=\frac{1}{4\pi \left(8.85\times {10}^{-12}\text{F/m}\right)}\frac{\left(+0.06\text{C}\right)\left(+0.14\text{C}\right)}{{\left(2.5\text{m}\right)}^2}\\ =1.21\times {10}^7\text{N}\end{array}$

Conclusion:

The magnitude of the force on $0.06\text{C}$ charge by $0.14\text{C}$ is $1.21\times {10}^7\text{N}$.

To determine

The magnitude of the force on $0.06\text{C}$ charge by $0.09\text{C}$.

Answer to Problem 1SP

The magnitude of the force on $0.06\text{C}$ charge by $0.09\text{C}$ is $2.16\times {10}^7\text{N}$.

Explanation of Solution

Given Info: The distance between the charges at C and B is $1.5\text{m}$, the charge at point C is $0.09\text{C}$, the charge at point B is $+0.06\text{C}$.

The diagram showing the charges

Write the expression to find the electrostatic force.

$F=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r^2}$

Here,

$F$ is the electrostatic force

$q_1$ is the first charge

$q_2$ is the second charge

$r$ is the distance between the charges

${\epsilon }_0$ is the permittivity of free space

Substitute $8.85\times {10}^{-12}\text{F/m}$ for ${\epsilon }_0$, $+0.06\text{C}$ for $q_1$, $0.09\text{C}$ for $q_2$, $1.5\text{m}$ for $r$ to find the electrostatic force.

$\begin{array}{c}F=\frac{1}{4\pi \left(8.85\times {10}^{-12}\text{F/m}\right)}\frac{\left(+0.06\text{C}\right)\left(0.09\text{C}\right)}{{\left(1.5\text{m}\right)}^2}\\ =2.16\times {10}^7\text{N}\end{array}$

Conclusion:

The magnitude of the force on $0.06\text{C}$ charge by $0.09\text{C}$ is $2.16\times {10}^7\text{N}$.

To determine

The net force exerted on the $0.06\text{C}$ charge by the other two charges.

Answer to Problem 1SP

The net force exerted on the $0.06\text{C}$ charge by the other two charges is $9.5\times {10}^6\text{N}$ to the left direction.

Explanation of Solution

Given Info: The magnitude of the force on $0.06\text{C}$ charge by $0.09\text{C}$ is $2.16\times {10}^7\text{N}$, the magnitude of the force on $0.06\text{C}$ charge by $0.14\text{C}$ is $1.21\times {10}^7\text{N}$.

The diagram showing the charges

Write the expression to find the net electrostatic force on charge at B.

$F=F_A+F_C$

Here,

$F_A$ is the force exerted on charge at B by the charge at A

$F_C$ is the force exerted on charge at B by the charge at C

F is the net electrostatic force on charge at B

Consider the right direction to be positive.

Substitute $1.21\times {10}^7\text{N}$ for $F_A$ and $-2.16\times {10}^7\text{N}$ for $F_C$ to find the electrostatic force.

$\begin{array}{c}F=\left(1.21\times {10}^7\text{N}\right)+\left(-2.16\times {10}^7\text{N}\right)\\ =-9.5\times {10}^6\text{N}\end{array}$

Since, the net force is negative; the direction of the net force will be in the left direction.

Conclusion:

The net force exerted on the $0.06\text{C}$ charge by the other two charges is $9.5\times {10}^6\text{N}$ to the left direction.

To determine

The direction and magnitude of the electric field at the point B.

Answer to Problem 1SP

The direction and magnitude of the electric field at point B is $1.58\times {10}^8\text{N/C}$ to the left direction.

Explanation of Solution

Given Info: The net force exerted on the $0.06\text{C}$ charge by the other two charges is $9.5\times {10}^6\text{N}$ to the left direction.

The diagram showing the charges

Write the expression to find the electric field.

$\overrightarrow{E}=\frac{\overrightarrow{F}}{q}$

Here,

$\overrightarrow{E}$ is the electric field

$\overrightarrow{F}$ is the force exerted on the charge.

q is the charge

Substitute $9.5\times {10}^6\text{N}$ for $\overrightarrow{F}$ and $0.06\text{C}$ for q to find the electric field at point B.

$\begin{array}{c}\overrightarrow{E}=\frac{9.5\times {10}^6\text{N}}{0.06\text{C}}\\ =1.58\times {10}^8\text{N/C}\end{array}$

The direction of the electric field is in the left direction.

Conclusion:

The direction and magnitude of the electric field at point B is $1.58\times {10}^8\text{N/C}$ to the left direction.

To determine

The direction and magnitude of the electrostatic force exerted on the new charge if the $0.06\text{C}$ charge is replaced by a charge of $-0.17\text{C}$.

Answer to Problem 1SP

The direction and magnitude of the electrostatic force exerted on the new charge if the $0.06\text{C}$ charge is replaced by a charge of $-0.17\text{C}$ is $2.69\times {10}^7\text{N}$ to the right direction.

Explanation of Solution

Given Info: The direction and magnitude of the electric field at point B is $1.58\times {10}^8\text{N/C}$ to the left direction.

The diagram showing the charges

Write the expression to find the electrostatic force.

$\overrightarrow{F}=q\overrightarrow{E}$

Here,

$\overrightarrow{E}$ is the electric field

$\overrightarrow{F}$ is the force exerted on the charge.

q is the charge

Substitute $1.58\times {10}^8\text{N/C}$ for $\overrightarrow{E}$ and $-0.17\text{C}$ for q to find the electrostatic force point B.

$\begin{array}{c}\overrightarrow{F}=\left(-0.17\text{C}\right)\left(1.58\times {10}^8\text{N/C}\right)\\ =-2.69\times {10}^7\text{N}\end{array}$

The direction of the electrostatic force will be in the right direction.

Conclusion:

The direction and magnitude of the electrostatic force exerted on the new charge if the $0.06\text{C}$ charge is replaced by a charge of $-0.17\text{C}$ is $2.69\times {10}^7\text{N}$ to the right direction.