Chapter 19, Problem 65P

(a)

To determine

The magnitude of total electric force exerted on the charge at the lower left corner due to the other three charges.

Answer to Problem 65P

The magnitude of total electric force exerted on the charge at the lower left corner due to the other three charges is $\underset{\_}{40.8\text{N}}$.

Explanation of Solution

Figure 1 represents four charges of equal magnitude of $10.0\mu \text{C}$.

Write the expression for angle $\theta$ from Figure 1,

    $\theta ={\tan }^{-1}\left(\frac{r_{\text{AB}}}{r_{\text{AD}}}\right)$        (I)

Here, $\theta$ is the angle, $r_{\text{AB}}$ is the distance between the charges A and B, and $r_{\text{AD}}$ is the distance between the charges A and D.

Write the expression for magnitude of electric force due to charge 1.

    $F_1=\frac{k_{\text{e}}{q}_1{q}_4}{r_{14}^2}$        (II)

Here, $F_1$ is the electric force, $k_{\text{e}}$ is the coulomb’s constant, $q_1$, $q_4$ are the charges, and $r_{14}$ is the distance between the charges $q_1$ and $q_4$.

Write the expression for magnitude of electric force due to charge 2.

    $F_2=\frac{k_{\text{e}}{q}_2{q}_4}{r_{24}^2}$        (III)

Here, $F_2$ is the electric force, $k_{\text{e}}$ is the coulomb’s constant, $q_2$, $q_4$ are the charges, and $r_{24}$ is the distance between the charges $q_2$, and $q_4$.

Write the expression for magnitude of electric force due to charge 1.

    $F_3=\frac{k_{\text{e}}{q}_3{q}_4}{r_{34}^2}$        (IV)

Here, $F_3$ is the electric force, $k_{\text{e}}$ is the coulomb’s constant, $q_2$, $q_4$ are the charges, and $r_{34}$ is the distance between the charges $q_2$, and $q_4$.

Write the expression for horizontal component of the resultant force.

    $F_{\text{x}}=-F_3-F_2\cos \theta$        (V)

Write the expression for vertical component of the resultant force.

    $F_y=-F_1-F_2\sin \theta$        (VI)

Write the expression for the resultant force.

    $F_{\text{net}}=\sqrt{F_{\text{x}}^2+F_{\text{y}}^2}$        (VII)

Conclusion:

Substitute $15.0\text{m}$ for $r_{\text{AB}}$, and $60.0\text{m}$ for $r_{\text{AD}}$ in equation (I), to find $\theta$.

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{15.0\text{m}}{60.0\text{m}}\right)\\ =14.0°\end{array}$

Substitute $8.99\times {10}^9\text{N}{\text{.m}}^2/{\text{C}}^2$ for $k_{\text{e}}$, $10.0\text{μC}$ for $q_1$, $10.0\text{μC}$ for $q_4$, and $0.150\text{m}$ for $r_{\text{AB}}$, in equation (II) to find $F_1$.

    $\begin{array}{c}{F}_1=\frac{\left(8.99\times {10}^9\text{N}{\text{.m}}^2/{\text{C}}^2\right)\left(10.0\text{μC}\times \frac{1\text{C}}{{10}^6\text{μC}}\right)\left(10.0\text{μC}\times \frac{1\text{C}}{{10}^6\text{μC}}\right)}{{\left(0.150\text{m}\right)}^2}\\ =40.0\text{N}\end{array}$

Substitute $8.99\times {10}^9\text{N}{\text{.m}}^2/{\text{C}}^2$ for $k_{\text{e}}$, $10.0\text{μC}$ for $q_2$, $10.0\text{μC}$ for $q_4$, and $0.618\text{m}$ for $r_{\text{AB}}$, in equation (III) to find $F_2$.

    $\begin{array}{c}{F}_2=\frac{\left(8.99\times {10}^9\text{N}{\text{.m}}^2/{\text{C}}^2\right)\left(10.0\text{μC}\times \frac{1\text{C}}{{10}^6\text{μC}}\right)\left(10.0\text{μC}\times \frac{1\text{C}}{{10}^6\text{μC}}\right)}{{\left(0.618\text{m}\right)}^2}\\ =2.35\text{N}\end{array}$

Substitute $8.99\times {10}^9\text{N}{\text{.m}}^2/{\text{C}}^2$ for $k_{\text{e}}$, $10.0\text{μC}$ for $q_3$, $10.0\text{μC}$ for $q_4$, and $0.600\text{m}$ for $r_{\text{AB}}$, in equation (IV) to find $F_3$.

    $\begin{array}{c}{F}_3=\frac{\left(8.99\times {10}^9\text{N}{\text{.m}}^2/{\text{C}}^2\right)\left(10.0\text{μC}\times \frac{1\text{C}}{{10}^6\text{μC}}\right)\left(10.0\text{μC}\times \frac{1\text{C}}{{10}^6\text{μC}}\right)}{{\left(0.600\text{m}\right)}^2}\\ =2.50\text{N}\end{array}$

Substitute $2.50\text{N}$ for $F_3$, $2.35\text{N}$ for $F_2$, and $14.0°$ for $\theta$ in equation (V), to find $F_{\text{x}}$.

    $\begin{array}{c}{F}_{\text{x}}=-2.50\text{N}-2.35\text{N}\cos 14.0°\\ =-4.78\text{N}\end{array}$

Substitute $40.0\text{N}$ for $F_1$, $2.35\text{N}$ for $F_2$, and $14.0°$ for $\theta$ in equation (VI), to find $F_{\text{y}}$.

    $\begin{array}{c}{F}_{\text{y}}=-40.0\text{N}-2.35\text{Nsin}14.0°\\ =-40.5\text{N}\end{array}$

Substitute $-4.78\text{N}$ for $F_{\text{x}}$, and $-40.5\text{N}$ for $F_{\text{y}}$ in equation (VII), to find $F_{\text{net}}$.

    $\begin{array}{c}{F}_{\text{net}}=\sqrt{{\left(-4.78\text{N}\right)}^2+{\left(-40.5\text{N}\right)}^2}\\ =40.8\text{N}\end{array}$

Therefore, the magnitude of total electric force exerted on the charge at the lower left corner due to the other three charges is $\underset{\_}{40.8\text{N}}$.

(b)

To determine

The direction of total electric force exerted on the charge at the lower left corner due to the other three charges.

Answer to Problem 65P

The direction of total electric force exerted on the charge at the lower left corner due to the other three charges is $\underset{\_}{263°}$.

Explanation of Solution

Write the expression for angle between the vertical component and the horizontal component of the resultant electric force from Figure 1.

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

Here, $\phi$ is the angle between the vertical component and the horizontal component of the resultant electric force.

Conclusion:

Substitute $-4.78\text{N}$ for $F_{\text{x}}$, and $-40.5\text{N}$ for $F_{\text{y}}$ in equation (VIII), to find $\phi$.

    $\begin{array}{c}\phi ={\tan }^{-1}\left(\frac{-40.5\text{N}}{-4.78\text{N}}\right)\\ =263°\end{array}$

Therefore, the direction of total electric force exerted on the charge at the lower left corner due to the other three charges is $\underset{\_}{263°}$.