Chapter 22, Problem 12P

(a)

To determine

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

Answer to Problem 12P

The $x-$ component of the electric force exerted by $A$ on $C$ is $0$.

Explanation of Solution

Given info: The charge of particle $A$ is $3.00\times {10}^{-4}\text{C}$, charge of particle $B$ is $-6.00\times {10}^{-4}\text{C}$, and charge of particle $C$ is $1.00\times {10}^{-4}\text{C}$, location of particle $A$ is origin, location of particle $B$ is $\left(4.00\text{m,}\text{0}\right)$ and  location of particle $C$ is $\left(0,3.00\text{m}\right)$.

The diagram for the given condition is shown below.

Figure 1

The formula to calculate the electrical force is,

$F_{{\text{AC}}_{\text{x}}}=\frac{k_e{q}_1{q}_2}{r^2}$

Here,

$k_e$ is the constant.

$q_1$ is the charge on particle $A$.

$q_2$ is the charge on particle $C$.

$r$ is the distance between the particles $A$ and $C$ on $x-$ axis.

The particle $A$ is at origin that is the abscissa and the ordinate both are $0$.and the particle $C$ is at the distance of $3.00\text{m}$ from the $y-$ axis and the abscissa is $0$.

The distance from the $x-$ axis is $0$ for both the particles $A$ and $C$. So, no net force will act on the particle $A$ by $C$.

Thus, the $x$ component of the electric force exerted by $A$ on $C$ is,

$F_{{\text{AC}}_{\text{x}}}=0$

Conclusion:

Therefore, the $x$ component of the electric force exerted by $A$ on $C$ is $0$.

(b)

To determine

The $y-$ component of the force exerted by $A$ on $C$.

Answer to Problem 12P

The $y-$ component of the force exerted by $A$ on $C$ is $30\text{N}$.

Explanation of Solution

Given info: The charge of particle $A$ is $3.00\times {10}^{-4}\text{C}$, charge of particle $B$ is $-6.00\times {10}^{-4}\text{C}$, and charge of particle $C$ is $1.00\times {10}^{-4}\text{C}$, location of particle $A$ is origin, location of particle $B$ is $\left(4.00\text{m,}\text{0}\right)$ and  location of particle $C$ is $\left(0,3.00\text{m}\right)$.

The formula to calculate the electrical force is,

$F_{{\text{AC}}_{\text{y}}}=\frac{k_e{q}_1{q}_2}{r^2}$

Substitute $3.00\times {10}^{-4}\text{C}$ for $q_1$, $1.00\times {10}^{-4}\text{C}$ for $q_2$, $3.00\text{m}$ to calculate $F_{{\text{AC}}_{\text{y}}}$ as,

$\begin{array}{c}{F}_{{\text{AC}}_{\text{y}}}=\frac{\left(9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\left(3.00\times {10}^{-4}\text{C}\right)\left(1.00\times {10}^{-4}\text{C}\right)}{{\left(3.00\text{m}\right)}^2}\\ =30\text{N}\end{array}$

Conclusion:

Therefore, the $y-$ component of the force exerted by $A$ on $C$ is $30\text{N}$.

(c)

To determine

The magnitude of the force exerted by $B$ on $C$.

Answer to Problem 12P

The magnitude of the force exerted by $B$ on $C$ is $21.6\text{N}$.

Explanation of Solution

Given info: The charge of particle $A$ is $3.00\times {10}^{-4}\text{C}$, charge of particle $B$ is $-6.00\times {10}^{-4}\text{C}$, and charge of particle $C$ is $1.00\times {10}^{-4}\text{C}$, location of particle $A$ is origin, location of particle $B$ is $\left(4.00\text{m,}\text{0}\right)$ and  location of particle $C$ is $\left(0,3.00\text{m}\right)$.

By Pythagoras theorem the distance between $B$ and $C$ is,

$\begin{array}{c}{\left(BC\right)}^2={\left(AC\right)}^2+{\left(AB\right)}^2\\ ={\left(3\text{m}\right)}^2+{\left(4\text{m}\right)}^2\\ BC=\sqrt{\left(16+9\right){\text{m}}^{\text{2}}}\\ =5\text{m}\end{array}$

Thus, the distance between $B$ and $C$ is $5\text{m}$.

The formula to calculate the electrical force is,

$F_{BC}=\frac{k_e{q}_1{q}_2}{r^2}$

Here,

$k_e$ is the constant.

$q_1$ is the charge on particle $B$.

$q_2$ is the charge on particle $C$.

$r$ is the distance between the particles $B$ and $C$.

Substitute $-6.00\times {10}^{-4}\text{C}$ for $q_1$, $1.00\times {10}^{-4}\text{C}$ for $q_2$, $5.00\text{m}$ to find $F_{BC}$ as,

$\begin{array}{c}{F}_{BC}=\frac{k_e{q}_1{q}_2}{r^2}\\ =\frac{\left(9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\left(-6.00\times {10}^{-4}\text{C}\right)\left(1.00\times {10}^{-4}\text{C}\right)}{{\left(5.00\text{m}\right)}^2}\\ =-21.6\text{N}\\ \left|F_{\text{BC}}\right|=21.6\text{N}\end{array}$

The magnitude of the force exerted by $B$ on $C$ is $21.6\text{N}$.

Conclusion:

Therefore, the magnitude of the force exerted by $B$ on $C$ $21.6\text{N}$.

(d)

To determine

The $x-$ component of the force exerted by $B$ on $C$.

Answer to Problem 12P

The $x-$ component of the force exerted by $B$ on $C$ is $17.28\text{N}$.

Explanation of Solution

Given info: The charge of particle $A$ is $3.00\times {10}^{-4}\text{C}$, charge of particle $B$ is $-6.00\times {10}^{-4}\text{C}$, and charge of particle $C$ is $1.00\times {10}^{-4}\text{C}$, location of particle $A$ is origin, location of particle $B$ is $\left(4.00\text{m,}\text{0}\right)$ and  location of particle $C$ is $\left(0,3.00\text{m}\right)$.

From part (c), the magnitude of the force exerted by $B$ on $C$ is $21.6\text{N}$.

Resolve the side $BC$ into horizontal and vertical components as-The horizontal component is $\left|F_{BC}\right|\cos \alpha$ and vertical component is $\left|F_{\text{BC}}\right|\sin \text{}\alpha$

From Figure I

$\cos \alpha =\frac{4}{5}$

The formula to calculate the $x-$ component of force by $B$ on $C$ is,

$F_{\text{xBC}}=\left|F_{\text{BC}}\right|\cos \alpha$

Here,

$\left|F_{\text{BC}}\right|$ is the force exerted by $B$ on $C$

$\cos \alpha$ is the horizontal component of $BC$.

Substitute $21.6\text{N}$ for $\left|F_{\text{BC}}\right|$ and $\frac{4}{5}$ for $\cos \alpha$ to calculate $F_{\text{xBC}}$ as,

$\begin{array}{c}{F}_{\text{xBC}}=\left|F_{\text{BC}}\right|\cos \alpha \\ =\left(21.6\text{}\text{N}\right)\left(\frac{4}{5}\right)\\ =17.28\text{N}\end{array}$

Conclusion:

Therefore, the $x-$ component of the force exerted by $B$ on $C$ is $17.28\text{N}$.

(e)

To determine

The $y-$ component of the force exerted by $B$ on $C$.

Answer to Problem 12P

The $y-$ component of the force exerted by $B$ on $C$ is $12.96\text{N}$.

Explanation of Solution

Given info: The charge of particle $A$ is $3.00\times {10}^{-4}\text{C}$, charge of particle $B$ is $-6.00\times {10}^{-4}\text{C}$, and charge of particle $C$ is $1.00\times {10}^{-4}\text{C}$, location of particle $A$ is origin, location of particle $B$ is $\left(4.00\text{m,}\text{0}\right)$ and  location of particle $C$ is $\left(0,3.00\text{m}\right)$.

From part (c), the magnitude of the force exerted by $B$ on $C$ is $21.6\text{N}$.

Resolve the side $BC$ into horizontal and vertical components as-the horizontal component is $\left|F_{\text{BC}}\right|\cos \alpha$ and vertical component is $\left|F_{\text{BC}}\right|\sin \text{}\alpha$

From Figure I,

$\sin \alpha =\frac{3}{5}$

The formula to calculate the $y-$ component of force by $B$ $C$ is,

$F_{\text{yBC}}=\left|F_{\text{BC}}\right|\sin \alpha$

Here,

$\left|F_{BC}\right|$ is the force exerted by $B$ on $C$

$\sin \alpha$ is the vertical component of $BC$.

Substitute $21.6\text{N}$ for $\left|F_{\text{BC}}\right|$ and $\frac{3}{5}$ for $\sin \alpha$ in the above formula as,

$\begin{array}{c}{F}_{\text{yBC}}=\left|F_{\text{BC}}\right|\sin \alpha \\ =\left(21.6\text{}\text{N}\right)\left(\frac{3}{5}\right)\\ =12.96\text{N}\end{array}$

Conclusion:

Therefore, the $y-$ component of the force exerted by $B$ on $C$ is $12.96\text{N}$.

(f)

To determine

The resultant $x-$ component of the electric force acting on $C$.

Answer to Problem 12P

The resultant $x-$ component of the electric force acting on $C$ is $17.28\text{N}$

Explanation of Solution

Given info: The charge of particle $A$ is $3.00\times {10}^{-4}\text{C}$, charge of particle $B$ is $-6.00\times {10}^{-4}\text{C}$, and charge of particle $C$ is $1.00\times {10}^{-4}\text{C}$, location of particle $A$ is origin, location of particle $B$ is $\left(4.00\text{m,}\text{0}\right)$ and  location of particle $C$ is $\left(0,3.00\text{m}\right)$.

From part (a), the $x-$ component of the force exerted by $A$ on $C$ is,

$\left|F_{\text{xAC}}\right|=0$

From part (d), the $x-$ component of the force exerted by $B$ on $C$ is,

$\left|F_{\text{xBC}}\right|=17.28\text{N}$

The formula to calculate the resultant force acting on the particle $C$ is,

$F_{\text{xC}}=F_{\text{xAC}}+F_{\text{xBC}}$

Here,

$F_{\text{xC}}$ is the sum of $x-$ component of the force acting on $C$.

$F_{\text{xAC}}$ is the $x-$ component of the force exerted by $A$ on $C$.

$F_{\text{xBC}}$ is the $x-$ component of the force exerted by $B$ on $C$.

Substitute $0$ for $\left|F_{\text{xAC}}\right|$ and $17.28\text{N}$ for $\left|F_{\text{xBC}}\right|$ to calculate $F_{\text{xC}}$ as,

$\begin{array}{c}{F}_{\text{xC}}=F_{\text{xAC}}+F_{\text{xBC}}\\ =0+17.28\text{N}\\ \text{=17}\text{.28}\text{N}\end{array}$

Conclusion:

Therefore, the resultant $x-$ component of the electric force acting on $C$ is $17.28\text{N}$.

(g)

To determine

The resultant $y-$ component of the electric force acting on $C$.

Answer to Problem 12P

The resultant $y-$ component of the electric force acting on $C$ is $17.28\text{N}$

Explanation of Solution

Given info: The charge of particle $A$ is $3.00\times {10}^{-4}\text{C}$, charge of particle $B$ is $-6.00\times {10}^{-4}\text{C}$, and charge of particle $C$ is $1.00\times {10}^{-4}\text{C}$, location of particle $A$ is origin, location of particle $B$ is $\left(4.00\text{m,}\text{0}\right)$ and  location of particle $C$ is $\left(0,3.00\text{m}\right)$.

From part (b), the $y-$ component of the force exerted by $A$ on $C$ is,

$\left|F_{\text{yAC}}\right|=30\text{N}$

From part (e), the $y-$ component of the force exerted by $B$ on $C$ is,

$\left|F_{\text{yBC}}\right|=17.28\text{N}$

The formula to calculate the resultant force acting on the particle $C$ is,

$F_{\text{yC}}=F_{\text{yAC}}+F_{\text{yBC}}$

Here,

$F_{\text{yC}}$ is the sum of $y-$ component of the force acting on $C$.

$F_{\text{yAC}}$ is the $y-$ component of the force exerted by $A$ on $C$.

$F_{\text{yBC}}$ is the $y-$ component of the force exerted by $B$ on $C$.

Substitute $30\text{N}$ for $\left|F_{\text{yAC}}\right|$ and $12.96\text{N}$ for $\left|F_{\text{yBC}}\right|$ to calculate $F_{\text{yC}}$ as,

$\begin{array}{c}{F}_{\text{yC}}=F_{\text{yAC}}+F_{\text{yBC}}\\ =30\text{N}+12.96\text{N}\\ \text{=42}\text{.96}\text{N}\end{array}$

Conclusion:

Therefore, the resultant $y-$ component of the electric force acting on $C$ is $42.96\text{N}$.

(h)

To determine

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

Answer to Problem 12P

The magnitude and direction of the resultant electric force acting on $C$ is $46.30\text{N}$.

Explanation of Solution

Given info: The charge of particle $A$ is $3.00\times {10}^{-4}\text{C}$, charge of particle $B$ is $-6.00\times {10}^{-4}\text{C}$, and charge of particle $C$ is $1.00\times {10}^{-4}\text{C}$, location of particle $A$ is origin, location of particle $B$ is $\left(4.00\text{m,}\text{0}\right)$ and  location of particle $C$ is $\left(0,3.00\text{m}\right)$.

From part (g), the resultant $y-$ component of the electric force acting on $C$ is $42.96\text{N}$.

$\left|F_{\text{yC}}\right|=42.96\text{N}$

From part (f), the resultant $x-$ component of the electric force acting on $C$ is

$\left|F_{\text{xC}}\right|=17.28\text{N}$

The formula to calculate the resultant force acting on the particle $C$ is,

$F_{\text{C}}=\sqrt{F^2{}_{\text{yC}}+F^2{}_{\text{xC}}}$

Here,

$F_{\text{C}}$ is the resultant force acting on $C$.

$F_{\text{yC}}$ is the resultant $y-$ component of the force exerted by $A$ on $C$.

$F_{\text{xC}}$ is the $x-$ component of the force exerted by $B$ on $C$.

Substitute $42.96\text{N}$ for $\left|F_{\text{yC}}\right|$ and $17.28\text{N}$ for $\left|F_{\text{xC}}\right|$ to calculate $F_{\text{C}}$  as,

$\begin{array}{c}{F}_{\text{C}}=\sqrt{F^2{}_{\text{yC}}+F^2{}_{\text{xC}}}\\ =\sqrt{{\left(42.96\text{N}\right)}^2+{\left(17.28\text{N}\right)}^2}\\ =\sqrt{1845.56\text{}\text{N}+298.59\text{N}}\\ =46.30\text{N}\end{array}$

The formula to calculate the direction of the resultant force acting on $C$ is,

$\theta ={\tan }^{-1}\left(\frac{F_{\text{yC}}}{F_{\text{xC}}}\right)$

Here,

$F_{\text{y}}$ is the resultant $y-$ component force acting on the particle $C$

$F_{\text{x}}$ is the resultant $x-$ component force acting on the particle $C$.

Substitute $42.96\text{N}$ for $F_{\text{yC}}$ and $17.28\text{N}$ for $F_{\text{xC}}$ to calculate $\theta$ is,

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{F_{yC}}{F_{xC}}\right)\\ ={\tan }^{-1}\left(\frac{42.96\text{N}}{17.28\text{N}}\right)\\ =5.131°\end{array}$

The direction of the resultant force is counterclockwise from $+x$ axis.

Conclusion:

Therefore, the magnitude and direction of the resultant electric force acting on $C$ is $46.30\text{N}$.