Chapter 19, Problem 10P

Particle A of charge 3.00 × 10^–4 C is at the origin, particle B of charge –6.00 × 10^–1 C is at (4.00 m, 0), and particle C of charge 1.00 × 10^–4 C is at (0, 3.00 in). We wish to find the net electric force on C. (a) What is the x component of the electric force exerted by A on C? (b) What is the y component of the force exerted by A on C? (c) Kind the magnitude of the force exerted by B on C. (d) Calculate the x component of the force exerted by B on C. (e) Calculate the y component of the force exerted by B on C. (f) Sum the two x components from parts (a) and (d) to obtain the resultant x component of the electric force acting on C. (g) Similarly, find the y component of the resultant force vector acting on C. (h) Kind the magnitude and direction of the resultant electric force acting on C.

To determine

The $x$ component of electric force exerted by charge A on charge C.

Answer to Problem 10P

The $x$ component of electric force exerted by charge A on charge is zero.

Explanation of Solution

Figure 1 represents the three charges A, B, and C, and the three charges are arranged in such a way that charge C is attracted by charge B, and repelled by charge A.

The distance between charge A and C is $3.00\text{m}$, and the distance between charge A and B is $4.00\text{m}$.

Write the distance between charge C and B from Figure 1.

    $\begin{array}{c}{r}_{\text{BC}}=\sqrt{{\left(4.00\text{m}\right)}^2+{\left(3.00\text{m}\right)}^2}\\ =5.00\text{m}\end{array}$        (I)

Write the angle between the horizontal and $r_{\text{BC}}$ from Figure 1.

    $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{3.00\text{m}}{4.00\text{m}}\right)\\ =36.9°\end{array}$        (II)

Write the expression for component of electric force exerted by charge A on charge C.

    $F_{\text{AC}}=k_{\text{e}}\frac{\left|q_{\text{A}}\right|\left|q_{\text{C}}\right|}{r_{\text{AC}}^2}$        (III)

Here, $F_{\text{AC}}$ is the electric force exerted by charge A on charge C, $q_{\text{A}}$ is the charge A, and $q_{\text{C}}$ is the charge C, $k_{\text{e}}$ is the coulomb’s constant, and $r_{\text{AC}}$ is the distance between charge C and A.

The $x$ component of electric force exerted by charge A on charge is zero, since cosine of $90°$ is zero from Figure 1.

Conclusion:

Therefore, the $x$ component of electric force exerted by charge A on charge is zero.

To determine

The $y$ component of electric force exerted by charge A on charge C.

Answer to Problem 10P

The $y$ component of electric force exerted by charge A on charge C is $\underset{\_}{30.0\text{N}}$

Explanation of Solution

The $y$ component of electric force exerted by charge A on charge C is equal to the electric force exerted by charge A on charge C from Figure 1.

Write the expression for $y$ component of electric force exerted by charge A on charge C.

    $F_{\text{AC}}=k_{\text{e}}\frac{\left|q_{\text{A}}\right|\left|q_{\text{C}}\right|}{r_{\text{AC}}^2}$        (IV)

Here, $F_{\text{AC}}$ is the electric force exerted by charge A on charge C, $q_{\text{A}}$ is the charge A, and $q_{\text{C}}$ is the charge C, $k_{\text{e}}$ is the coulomb’s constant, and $r_{\text{AC}}$ is the distance between charge C and A.

Conclusion:

Substitute $8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $k_{\text{e}}$, $3.00\times {10}^{-4}\text{C}$ for $\left|q_{\text{A}}\right|$, $1.00\times {10}^{-4}\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $\left|q_{\text{C}}\right|$, and $3.00\text{m}$ for $r_{\text{AC}}$ in equation (IV), to find $F_{\text{AC}}$.

    $\begin{array}{c}{F}_{\text{AC}}=\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\frac{\left|3.00\times {10}^{-4}\text{C}\right|\left|1.00\times {10}^{-4}\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right|}{{\left(3.00\text{m}\right)}^2}\\ =30.0\text{N}\end{array}$

Therefore, the $y$ component of electric force exerted by charge A on charge C is $\underset{\_}{30.0\text{N}}$.

To determine

The magnitude of electric force exerted by charge B on charge C.

Answer to Problem 10P

The magnitude of electric force exerted by charge B on charge is $\underset{\_}{21.6\text{N}}$.

Explanation of Solution

Write the expression for electric force exerted by charge B on charge C.

    $\left|F_{\text{BC}}\right|=k_{\text{e}}\frac{\left|q_{\text{B}}\right|\left|q_{\text{C}}\right|}{r_{\text{BC}}^2}$        (V)

Here, $F_{\text{BC}}$ is the electric force exerted by charge B on charge C, $q_{\text{B}}$ is the charge B, and $q_{\text{C}}$ is the charge C, $k_{\text{e}}$ is the coulomb’s constant, and $r_{\text{BC}}$ is the distance between charge C and B.

Conclusion:

Substitute $8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $k_{\text{e}}$, $6.00\times {10}^{-4}\text{C}$ for $\left|q_{\text{B}}\right|$, $1.00\times {10}^{-4}\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $\left|q_{\text{C}}\right|$, and $5.00\text{m}$ for $r_{\text{BC}}$ in equation (V), to find $F_{\text{AC}}$.

    $\begin{array}{c}{F}_{\text{AC}}=\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\frac{\left|6.00\times {10}^{-4}\text{C}\right|\left|1.00\times {10}^{-4}\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right|}{{\left(5.00\text{m}\right)}^2}\\ =21.6\text{N}\end{array}$

Therefore, the magnitude of electric force exerted by charge B on charge is $\underset{\_}{21.6\text{N}}$.

To determine

The $x$ component of electric force exerted by charge B on charge C.

Answer to Problem 10P

The $x$ component of electric force exerted by charge B on charge C is $\underset{\_}{17.3\text{N}}$.

Explanation of Solution

From subpart (b) the magnitude of force exerted by charge B on charge C is $\left|F_{\text{BC}}\right|=21.6\text{N}$.

Write the expression for $x$ component of electric force exerted by charge B on charge C.

    ${\left(F_{\text{BC}}\right)}_{\text{x}}=\left|F_{\text{BC}}\right|\cos \theta$        (VI)

Here, ${\left(F_{\text{BC}}\right)}_{\text{x}}$ is the $x$ component of electric force exerted by charge B on charge C, and $\left|F_{\text{BC}}\right|$ is the magnitude of force exerted by charge B on charge C.

Conclusion:

Substitute $21.6\text{N}$ for $\left|F_{\text{BC}}\right|$, and $36.9°$ for $\theta$ in equation (VI), to find ${\left(F_{\text{BC}}\right)}_{\text{x}}$.

    $\begin{array}{c}{\left(F_{\text{BC}}\right)}_{\text{x}}=\left|21.6\text{N}\right|\cos 36.9°\\ =17.3\text{N}\end{array}$

Therefore, the $x$ component of electric force exerted by charge B on charge C is $\underset{\_}{17.3\text{N}}$.

To determine

The $y$ component of electric force exerted by charge B on charge C.

Answer to Problem 10P

The $y$ component of electric force exerted by charge B on charge C is $\underset{\_}{-13.0\text{N}}$.

Explanation of Solution

From subpart (b) the magnitude of force exerted by charge B on charge C is $\left|F_{\text{BC}}\right|=21.6\text{N}$.

Write the expression for $y$ component of electric force exerted by charge B on charge C.

    ${\left(F_{\text{BC}}\right)}_{\text{x}}=-\left|F_{\text{BC}}\right|\sin \theta$        (VII)

Here, ${\left(F_{\text{BC}}\right)}_{\text{x}}$ is the $x$ component of electric force exerted by charge B on charge C, and $\left|F_{\text{BC}}\right|$ is the magnitude of force exerted by charge B on charge C.

Conclusion:

Substitute $21.6\text{N}$ for $\left|F_{\text{BC}}\right|$, and $36.9°$ for $\theta$ in equation (VII), to find ${\left(F_{\text{BC}}\right)}_{\text{y}}$.

    $\begin{array}{c}{\left(F_{\text{BC}}\right)}_{\text{x}}=-\left|21.6\text{N}\right|\sin 36.9°\\ =-13.0\text{N}\end{array}$

Therefore, the $y$ component of electric force exerted by charge B on charge C is $\underset{\_}{-13.0\text{N}}$.

To determine

The sum of $x$ component of electric force exerted by charge A on charge C and electric force exerted by charge B on charge C.

Answer to Problem 10P

The sum of $x$ component of electric force exerted by charge A on charge C, and charge B on charge C is $\underset{\_}{17.3\text{N}}$.

Explanation of Solution

Write the expression for sum of $x$ component of electric force exerted by charge A on charge C, and charge B on charge C.

    ${\left(F_{\text{R}}\right)}_{\text{x}}={\left(F_{\text{AC}}\right)}_{\text{x}}+{\left(F_{\text{BC}}\right)}_{\text{x}}$        (VIII)

Here, ${\left(F_{\text{R}}\right)}_{\text{x}}$ is the sum of $x$ component of electric force exerted by charge A on charge C, and charge B on charge C, ${\left(F_{\text{AC}}\right)}_{\text{x}}$ is the electric force exerted by charge A on charge C, and ${\left(F_{\text{BC}}\right)}_{\text{x}}$ is the electric force exerted by charge B on charge C.

Conclusion:

Substitute $0\text{N}$ for ${\left(F_{\text{AC}}\right)}_{\text{x}}$, and $17.3\text{N}$ for ${\left(F_{\text{BC}}\right)}_{\text{x}}$ in equation (VIII), to find ${\left(F_{\text{R}}\right)}_{\text{x}}$.

    $\begin{array}{c}{\left(F_{\text{R}}\right)}_{\text{x}}=0\text{N}+17.3\text{N}\\ \text{=17}\text{.3}\text{N}\end{array}$

Therefore, The sum of $x$ component of electric force exerted by charge A on charge C, and charge B on charge C is $\underset{\_}{17.3\text{N}}$.

To determine

The sum of $y$ component of electric force exerted by charge A on charge C and electric force exerted by charge B on charge C.

Answer to Problem 10P

The sum of $y$ component of electric force exerted by charge A on charge C and electric force exerted by charge B on charge C is $\underset{\_}{17.0\text{N}}$.

Explanation of Solution

Write the expression for sum of $y$ component of electric force exerted by charge A on charge C, and charge B on charge C.

    ${\left(F_{\text{R}}\right)}_{\text{y}}={\left(F_{\text{AC}}\right)}_{\text{y}}+{\left(F_{\text{BC}}\right)}_{\text{y}}$        (IX)

Here, ${\left(F_{\text{R}}\right)}_{\text{y}}$ is the sum of $y$ component of electric force exerted by charge A on charge C, and charge B on charge C, ${\left(F_{\text{AC}}\right)}_{\text{y}}$ is the electric force exerted by charge A on charge C, and ${\left(F_{\text{BC}}\right)}_{\text{y}}$ is the electric force exerted by charge B on charge C.

Conclusion:

Substitute $30.0\text{N}$ for ${\left(F_{\text{AC}}\right)}_{\text{y}}$, and $13.0\text{N}$ for ${\left(F_{\text{BC}}\right)}_{\text{y}}$ in equation (IX), to find ${\left(F_{\text{R}}\right)}_{\text{y}}$.

    $\begin{array}{c}{\left(F_{\text{R}}\right)}_{\text{y}}=30.0\text{N}+13.0\text{N}\\ \text{=17}\text{.0}\text{N}\end{array}$

Therefore, the sum of $y$ component of electric force exerted by charge A on charge C and electric force exerted by charge B on charge C is $\underset{\_}{17.0\text{N}}$.

To determine

The magnitude and direction of the resultant electric force acting on charge C.

Answer to Problem 10P

The magnitude of the resultant electric force acting on charge C is $\underset{\_}{24.3\text{N}}$, and direction of the resultant electric force acting on charge C is $\underset{\_}{44.5°}$ above the positive $x$ direction.

Explanation of Solution

Write the expression for magnitude of resultant electric force acting on charge C.

    $F_{\text{R}}=\sqrt{{\left(F_{\text{R}}\right)}_{\text{x}}^2+{\left(F_{\text{R}}\right)}_{\text{y}}^2}$        (X)

Here, $F_{\text{R}}$ is the magnitude of resultant electric force acting on charge, ${\left(F_{\text{R}}\right)}_{\text{x}}$ is the resultant electric force acting on charge C along $x$ direction, and ${\left(F_{\text{R}}\right)}_{\text{y}}$ is the resultant electric force acting on charge C along $y$ direction.

Write the expression for direction of the resultant electric force acting on charge C.

    $\phi ={\tan }^{-1}\left(\frac{{\left(F_{\text{R}}\right)}_{\text{y}}}{{\left(F_{\text{R}}\right)}_{\text{x}}}\right)$        (XI)

Here, $\phi$ is the direction of resultant electric force acting on charge C.

Conclusion:

Substitute $17.3\text{N}$ for ${\left(F_{\text{R}}\right)}_{\text{x}}$, and $17.0\text{N}$ for ${\left(F_{\text{R}}\right)}_{\text{y}}$ in equation (X), to find $F_{\text{R}}$.

    $\begin{array}{c}{F}_{\text{R}}=\sqrt{{\left(17.3\text{N}\right)}^2+{\left(17.0\text{N}\right)}^2}\\ =24.3\text{N}\end{array}$

Substitute $17.3\text{N}$ for ${\left(F_{\text{R}}\right)}_{\text{x}}$, and $17.0\text{N}$ for ${\left(F_{\text{R}}\right)}_{\text{y}}$ in equation (XI), to find $\phi$.

    $\begin{array}{c}\phi ={\tan }^{-1}\left(\frac{17.0\text{N}}{17.3\text{N}}\right)\\ 44.5°\end{array}$

Therefore, The magnitude of the resultant electric force acting on charge C is $\underset{\_}{24.3\text{N}}$, and direction of the resultant electric force acting on charge C is $\underset{\_}{44.5°}$ above the $+x$ direction.