Chapter 15, Problem 16P

Panicle A of charge 3.00 × 10⁻⁴ C is at the origin, particle B of charge −6.00 × 10⁻⁴ C is at (4.00 m, 0), and panicle C of charge 1.00 × 10⁻⁴ C is at (0, 3.00 m). (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) Find 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 to obtain the resultant x-component of the electric force acting on C. (g) Repeat part (f) for the y-component. (h) Find the magnitude and direction of the resultant electric force acting on C.

To determine

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

Answer to Problem 16P

Solution: The x component of electric force exerted by A on C is zero.

Explanation of Solution

Given info: The charges are $q_A=3.00\times {10}^{-4}\text{C}$ and $q_B=-6.00\times {10}^{-4}\text{C}$. The charges are placed at origin and (4.00 m, 0) respectively. The charge $q_C=1.00\times {10}^{-4}\text{C}$ is placed at (0, 3.000 m).

The electric force is given in the diagram below,

Thex component of electric force exerted by A on C is zero. This is because the force exerted by A on C will be along y axis only.

Conclusion:

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

To determine

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

Answer to Problem 16P

Solution: The y component of electric force exerted by A on C is 30.0 N.

Explanation of Solution

Given info: The charges are $q_A=3.00\times {10}^{-4}\text{C}$ and $q_B=-6.00\times {10}^{-4}\text{C}$. The charges are placed at origin and (4.00 m, 0) respectively. The charge $q_C=1.00\times {10}^{-4}\text{C}$ is placed at (0, 3.000 m).

Formula to calculate the y component of electric force exerted by A on C is,

${\left(F_{AC}\right)}_y=\frac{k_e{q}_A{q}_C}{r_{AC}^2}$

• $k_e$ is the Coulomb constant.
• $r_{AC}$ is the distance between the charges A and C.

Substitute $8.99\times {10}^9\text{N}{\text{.m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k_e$, $3.00\times {10}^{-4}\text{C}$ for $q_A$, $1.00\times {10}^{-4}\text{C}$ for $q_C$, 3.00 m for $r_{AC}$.

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

Conclusion:

The y component of electric force exerted by A on C is 30.0 N.

To determine

The magnitude of electric force exerted by B on C.

Answer to Problem 16P

Solution: The magnitude of electric force exerted by B on C is 21.6 N.

Explanation of Solution

Given info: The charges are $q_A=3.00\times {10}^{-4}\text{C}$ and $q_B=-6.00\times {10}^{-4}\text{C}$. The charges are placed at origin and (4.00 m, 0) respectively. The charge $q_C=1.00\times {10}^{-4}\text{C}$ is placed at (0, 3.000 m).

Formula to calculate themagnitude of electric force exerted by B on C is,

$F_{BC}=\frac{k_e{q}_B{q}_C}{r_{BC}^2}$

• $k_e$ is the Coulomb constant.
• $r_{BC}$ is the distance between the charges A and C.

Substitute $8.99\times {10}^9\text{N}{\text{.m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k_e$, $1.00\times {10}^{-4}\text{C}$ for $q_C$, $6.00\times {10}^{-4}\text{C}$ for $q_B$, 5.00 m for $r_{BC}$.

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

Conclusion:

The magnitude of electric force exerted by B on C is 21.6 N.

To determine

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

Answer to Problem 16P

Solution: The x component of electric force exerted by B on C is 17.3 N.

Explanation of Solution

Given info: The charges are $q_A=3.00\times {10}^{-4}\text{C}$ and $q_B=-6.00\times {10}^{-4}\text{C}$. The charges are placed at origin and (4.00 m, 0) respectively. The charge $q_C=1.00\times {10}^{-4}\text{C}$ is placed at (0, 3.000 m).

Thex component of electric force exerted by B on C is,

${\left(F_{BC}\right)}_x=F_{BC}\cos \theta$

• $\theta$ is the angle between AB and BC.

The value of $\cos \theta$ is,

$\cos \theta =\frac{r_{AB}}{r_{BC}}$

From the above equations,

${\left(F_{BC}\right)}_x=F_{BC}\frac{r_{AB}}{r_{BC}}$

Substitute 21.6 N for $F_{BC}$, 4.00 m for $r_{AB}$ and 5.00 m for $r_{BC}$.

$\begin{array}{c}{\left(F_{BC}\right)}_x=\left(21.6\text{N}\right)\left(\frac{4.00\text{m}}{5.00\text{m}}\right)\\ =17.3\text{N}\end{array}$

Conclusion:

The x component of electric force exerted by B on C is 17.3 N.

To determine

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

Answer to Problem 16P

Solution: The y component of electric force exerted by B on C is -13.0 N.

Explanation of Solution

Given info: The charges are $q_A=3.00\times {10}^{-4}\text{C}$ and $q_B=-6.00\times {10}^{-4}\text{C}$. The charges are placed at origin and (4.00 m, 0) respectively. The charge $q_C=1.00\times {10}^{-4}\text{C}$ is placed at (0, 3.000 m).

They component of electric force exerted by B on C is,

${\left(F_{BC}\right)}_y=-F_{BC}\sin \theta$

The value of $\sin \theta$ is,

$\sin \theta =\frac{r_{AC}}{r_{BC}}$

From the above equations,

${\left(F_{BC}\right)}_y=-F_{BC}\frac{r_{AC}}{r_{BC}}$

Substitute 21.6 N for $F_{BC}$, 3.00 m for $r_{AC}$ and 5.00 m for $r_{BC}$.

$\begin{array}{c}{\left(F_{BC}\right)}_y=-\left(21.6\text{N}\right)\left(\frac{3.00\text{m}}{5.00\text{m}}\right)\\ =-13.0\text{N}\end{array}$

Conclusion:

The y component of electric force exerted by B on C is -13.0 N.

To determine

The resultant x component of electric force on C.

Answer to Problem 16P

Solution: The resultant x component of electric force on C is 17.3 N.

Explanation of Solution

Given info: The charges are $q_A=3.00\times {10}^{-4}\text{C}$ and $q_B=-6.00\times {10}^{-4}\text{C}$. The charges are placed at origin and (4.00 m, 0) respectively. The charge $q_C=1.00\times {10}^{-4}\text{C}$ is placed at (0, 3.000 m).

The resultant x component of electric force on C is,

${\left(F_R\right)}_x={\left(F_{AC}\right)}_x+{\left(F_{BC}\right)}_x$

Substitute 0 N for ${\left(F_{AC}\right)}_x$ and 17.3 N for ${\left(F_{BC}\right)}_x$.

$\begin{array}{c}{\left(F_R\right)}_x=\left(\text{0 N}\right)+\left(\text{17}\text{.3 N}\right)\\ =\text{17}\text{.3 N}\end{array}$

Conclusion:

The resultant x component of electric force on C is 17.3 N.

To determine

The resultant y component of electric force on C.

Answer to Problem 16P

Solution: The resultant y component of electric force on C is 17.0 N.

Explanation of Solution

Given info: The charges are $q_A=3.00\times {10}^{-4}\text{C}$ and $q_B=-6.00\times {10}^{-4}\text{C}$. The charges are placed at origin and (4.00 m, 0) respectively. The charge $q_C=1.00\times {10}^{-4}\text{C}$ is placed at (0, 3.000 m).

The resultant y component of electric force on C is,

${\left(F_R\right)}_y={\left(F_{AC}\right)}_y+{\left(F_{BC}\right)}_y$

Substitute 30.0 N for ${\left(F_{AC}\right)}_y$ and -13 N for ${\left(F_{BC}\right)}_y$.

$\begin{array}{c}{\left(F_R\right)}_y=\left(\text{30 N}\right)+\left(\text{-13 N}\right)\\ =\text{17}\text{.0 N}\end{array}$

Conclusion:

The resultant x component of electric force on C is 17.0 N.

To determine

The resultant electric force on C.

Answer to Problem 16P

Solution: The resultant electric force on C is 24.3 N directed at an angle of ${44.5}^o$ above the x axis.

Explanation of Solution

Given info: The charges are $q_A=3.00\times {10}^{-4}\text{C}$ and $q_B=-6.00\times {10}^{-4}\text{C}$. The charges are placed at origin and (4.00 m, 0) respectively. The charge $q_C=1.00\times {10}^{-4}\text{C}$ is placed at (0, 3.000 m).

The resultant of electric force on C is,

$F_R=\sqrt{{\left(F_R\right)}_x^2+{\left(F_R\right)}_y^2}$

Substitute 17.3 N for ${\left(F_R\right)}_x$ and 17.0 N for ${\left(F_R\right)}_y$.

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

The direction of resultant electric force on C is,

$\theta ={\tan }^{-1}\left(\frac{{\left(F_R\right)}_y}{{\left(F_R\right)}_x}\right)$

Substitute 17.3 N for ${\left(F_R\right)}_x$ and 17.0 N for ${\left(F_R\right)}_y$.

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

Conclusion:

The resultant electric force on C is 24.3 N directed at an angle of $44.5°$ above the x axis