Chapter 17, Problem 117P

(a)

To determine

The total force on the $2.50\text{μC}$ charge.

Answer to Problem 117P

The total force on the $2.50\text{μC}$ charge is $\underset{\_}{6.24\text{N}}$ at $\underset{\_}{16.1°}$ below the $+x$ axis.

Explanation of Solution

The following figure gives the diagram of the total force acting on the body.

Write the expression for the $x$ component of force.

    $F_x=F_{1x}+F_{2x}$                                                                                                            (I)

Here, $F_{1x}$ is the force between the charges $q_1$ and $q_2$ in the $x$ direction, and $F_{2x}$ is the force between the charges $q_1$ and $q_3$ in the $x$ direction.

Write the expression for the $y$ component of force.

    $F_y=F_{1y}+F_{2y}$                                                                                                         (II)

Here, $F_{1y}$ is the force between the charges $q_1$ and $q_2$ in the $y$ direction, and $F_{2y}$ is the force between the charges $q_1$ and $q_3$ in the $y$ direction.

Rewrite equation (I) using Coulomb’s force between two charges.

    $F_x=\frac{k\left|q_1\right|\left|q_2\right|}{d^2}\cos 60°+\frac{k\left|q_1\right|\left|q_3\right|}{d^2}\cos \left(-60°\right)$                                                            (III)

Here, $k$ is a constant, $q_1$ is the first charge, $q_2$ is the second charge, $q_3$ is the third charge and $d$ is the side of the equilateral triangle.

Rewrite equation (II) using Coulomb’s force between two charges.

    $F_y=\frac{k\left|q_1\right|\left|q_2\right|}{d^2}\sin 60°+\frac{k\left|q_1\right|\left|q_3\right|}{d^2}\sin \left(-60°\right)$                                                            (IV)

Write the expression to find the total magnitude of the force.

    $F=\sqrt{F_x^2+F_y^2}$                                                                                                         (V)

Write the expression to find the direction of the force.

    $\theta ={\tan }^{-1}\frac{F_y}{F_x}$                                                                                                            (VI)

Conclusion:

Substitute equation (III) and (IV) in equation (V) to find the magnitude of total force.

    $\begin{array}{c}F=\sqrt{F_x^2+F_y^2}\\ =\sqrt{{\left(\frac{k\left|q_1\right|\left|q_2\right|}{d^2}\cos 60°+\frac{k\left|q_1\right|\left|q_3\right|}{d^2}\cos \left(-60°\right)\right)}^2+{\left(\frac{k\left|q_1\right|\left|q_2\right|}{d^2}\sin 60°+\frac{k\left|q_1\right|\left|q_3\right|}{d^2}\sin \left(-60°\right)\right)}^2}\\ =\frac{k\left|q_1\right|}{d^2}\sqrt{{\left(\cos 60°\left(\left|q_2\right|+\left|q_3\right|\right)\right)}^2+{\left(\sin 60°\left(\left|q_2\right|-\left|q_3\right|\right)\right)}^2}\end{array}$

Substitute $8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $k$, $2.50\text{μC}$ for $q_1$, $0.150\text{m}$ for $d$, $5.00\text{μC}$ for $q_2$ and $7.00\text{μC}$ for $q_3$ in the above equation.

    $\begin{array}{c}F=\frac{\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\left|2.50\text{μC}\right|}{{\left(0.150\text{m}\right)}^2}\sqrt{{\left(\cos 60°\left(\left|5.00\text{μC}\right|+\left|7.00\text{μC}\right|\right)\right)}^2+{\left(\sin 60°\left(\left|5.00\text{μC}\right|-\left|7.00\text{μC}\right|\right)\right)}^2}\\ =6.24\text{N}\end{array}$

Substitute equation (III) and (IV) in equation (VI) to find the direction of total force.

    $\theta ={\tan }^{-1}\frac{\frac{k\left|q_1\right|\left|q_2\right|}{d^2}\sin 60°+\frac{k\left|q_1\right|\left|q_3\right|}{d^2}\sin \left(-60°\right)}{\frac{k\left|q_1\right|\left|q_2\right|}{d^2}\cos 60°+\frac{k\left|q_1\right|\left|q_3\right|}{d^2}\cos \left(-60°\right)}$

Substitute $2.50\text{μC}$ for $q_1$, $0.150\text{m}$ for $d$, $5.00\text{μC}$ for $q_2$ and $7.00\text{μC}$ for $q_3$ in the above equation.

    $\begin{array}{c}\theta ={\tan }^{-1}\frac{\frac{k\left|q_1\right|\left|q_2\right|}{d^2}\sin 60°+\frac{k\left|q_1\right|\left|q_3\right|}{d^2}\sin \left(-60°\right)}{\frac{k\left|q_1\right|\left|q_2\right|}{d^2}\cos 60°+\frac{k\left|q_1\right|\left|q_3\right|}{d^2}\cos \left(-60°\right)}\\ ={\tan }^{-1}\left(\frac{\sin 60°\left(\left|q_2\right|-\left|q_3\right|\right)}{\cos 60°\left(\left|q_2\right|+\left|q_3\right|\right)}\right)\\ =-16.1°\end{array}$

Therefore, the total force on the $2.50\text{μC}$ charge is $\underset{\_}{6.24\text{N}}$ at $\underset{\_}{16.1°}$ below the $+x$ axis.

(b)

To determine

The electric potential energy of the three charges.

Answer to Problem 117P

The electric potential energy of the three charges is $\underset{\_}{-2.40\text{J}}$.

Explanation of Solution

Write the expression for the total potential energy of the charges.

    $U=\frac{k{q}_1{q}_2}{d}+\frac{k{q}_1{q}_3}{d}+\frac{k{q}_2{q}_3}{d}$

Conclusion:

Substitute $8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $k$, $2.50\text{μC}$ for $q_1$, $0.150\text{m}$ for $d$, $5.00\text{μC}$ for $q_2$ and $-7.00\text{μC}$ for $q_3$ in the above equation.

    $\begin{array}{c}U=\frac{k{q}_1{q}_2}{d}+\frac{k{q}_1{q}_3}{d}+\frac{k{q}_2{q}_3}{d}\\ =\frac{k}{d}\left(q_1{q}_2+q_1{q}_3+q_2{q}_3\right)\\ =\frac{8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2}{0.150\text{m}}\left(\left(2.50\text{μC}\right)\left(5.00\text{μC}\right)+\left(2.50\text{μC}\right)\left(-7.00\text{μC}\right)+\left(5.00\text{μC}\right)\left(-7.00\text{μC}\right)\right)\\ =-2.40\text{J}\end{array}$

Therefore, the electric potential energy of the three charges is $\underset{\_}{-2.40\text{J}}$.