Chapter 23, Problem 71AP

To determine

The total force on the charge of $3.00\mu \text{C}$ at the center of curvature.

Answer to Problem 71AP

The total force on the charge of $3.00\mu \text{C}$ at the center of curvature is $-0.706\widehat{j}\text{N}$.

Explanation of Solution

Write the expression for the force on the line of positive charge.

    $F=q\overrightarrow{E}$                                                                                                                      (I)

Here, $F$ is the total charge, $q$ is the electric charge, $\overrightarrow{E}$ is the electric field.

Write the expression for the electric field.

    $E=\frac{kq}{r^2}$                                                                                                                      (II)

Here, $E$ is the electric field, $k$ is the constant, $q$ is the electric charge, $r$ is the distance between the charge and the position considered.

Write the expression for resolving the electric field in $x$ axis.

    $E_{\text{x}}=dE\sin \theta \widehat{i}$                                                                                                         (III)

Here, $E_{\text{x}}$ is the electric field along $x$ axis, $dE$ is the small change of the field is the angle of the along that direction, $\theta$ is the angle of the field.

Write the expression for resolving the electric field in $y$ direction.

    $E_{\text{y}}=dE\cos \theta \widehat{j}$                                                                                                        (IV)

Here, $E_{\text{y}}$ is the electric field along $y$ axis.

Write the expression for the small change in the charge over small change in length.

    $\begin{array}{c}dQ=\lambda dl\\ dQ=\lambda \left(Rd\theta \right)\end{array}$                                                                                                          (V)

Here, $dQ$ is the small change in charge, $\lambda$ is the charge per unit length, $dl$ is the small change in length, $R$ is the radius of curvature, $d\theta$ is the angle in the small change in the length.

Write the expression for the small change in charge.

    $dE=\frac{kdQ}{R^2}$                                                                                                               (VI)

Conclusion:

Substitute $\lambda \left(Rd\theta \right)$ for $dQ$ in the equation (VI) to find $dE$.

    $dE=\frac{k\lambda \left(Rd\theta \right)}{R^2}$                                                                                                      (VII)

Upon integration, the small change in field becomes the total electric field.

Integrate equation (VII).

    $\begin{array}{c}\overrightarrow{E}={\displaystyle \underset{-\pi /2}{\overset{\pi /2}{\int }}\frac{k\lambda \left(Rd\theta \right)}{R^2}\cos \theta }\left(-\widehat{j}\right)\\ =\frac{k}{R}{\displaystyle \underset{-\pi /2}{\overset{\pi /2}{\int }}\lambda \cos \theta }d\theta \left(-\widehat{j}\right)\end{array}$

Substitute ${\lambda }_{\text{ο}}\cos \theta$ for $\lambda$ in the above equation.

    $\begin{array}{c}\overrightarrow{E}=\frac{k}{R}{\displaystyle \underset{-\pi /2}{\overset{\pi /2}{\int }}\left({\lambda }_{\text{ο}}\cos \theta \right)\cos \theta }d\theta \left(-\widehat{j}\right)\\ =\frac{k{\lambda }_{\text{ο}}}{R}{\displaystyle \underset{-\pi /2}{\overset{\pi /2}{\int }}{\cos }^2\theta }d\theta \left(-\widehat{j}\right)\\ =\frac{k{\lambda }_{\text{ο}}}{R}{\displaystyle \underset{-\pi /2}{\overset{\pi /2}{\int }}\left[\frac{1}{2}\left(1+\cos 2\theta \right)\right]}d\theta \left(-\widehat{j}\right)\\ =-\frac{k{\lambda }_{\text{ο}}}{R}{\displaystyle \underset{-\pi /2}{\overset{\pi /2}{\int }}\left[\frac{1}{2}\left(1+\cos 2\theta \right)\right]}d\theta \left(\widehat{j}\right)\end{array}$

Calculate further.

    $\begin{array}{c}\overrightarrow{E}=-\frac{k{\lambda }_{\text{ο}}}{2R}{\displaystyle \underset{-\pi /2}{\overset{\pi /2}{\int }}\left(1+\cos 2\theta \right)}d\theta \widehat{j}\\ =-\frac{k{\lambda }_{\text{ο}}}{2R}{\left[\theta +\sin 2\theta \right]}_{-\pi /2}^{\pi /2}\widehat{j}\\ =-\frac{k{\lambda }_{\text{ο}}}{2R}\left[\pi +0\right]\widehat{j}\\ =-\frac{k\pi {\lambda }_{\text{ο}}}{2R}\widehat{j}\end{array}$                                                                             (VIII)

Integrate equation (V) to get the total charge.

    $\begin{array}{c}Q={\displaystyle \underset{-\pi /2}{\overset{\pi /2}{\int }}\lambda \left(Rd\theta \right)}\\ ={\displaystyle \underset{-\pi /2}{\overset{\pi /2}{\int }}{\lambda }_{\text{ο}}\cos \theta \left(Rd\theta \right)}\\ ={\lambda }_{\text{ο}}R{\displaystyle \underset{-\pi /2}{\overset{\pi /2}{\int }}\cos \theta d\theta }\\ ={\lambda }_{\text{ο}}R{\left[\sin \theta \right]}_{-\pi /2}^{\pi /2}\end{array}$

Calculate further.

    $\begin{array}{c}Q={\lambda }_{\text{ο}}R\left(1-\left(-1\right)\right)\\ ={\lambda }_{\text{ο}}R\left(1+1\right)\\ =2{\lambda }_{\text{ο}}R\end{array}$

Rewrite the above expression to get ${\lambda }_{\text{ο}}$.

    ${\lambda }_{\text{ο}}=\frac{Q}{2R}$                                                                                                                  (IX)

Substitute $12\mu \text{C}$ for $Q$ and $60.0\text{cm}$ for $R$ in the equation (IX) to find ${\lambda }_{\text{ο}}$.

    $\begin{array}{c}{\lambda }_{\text{ο}}=\frac{12\mu \text{C}}{2\times 60.0\text{cm}}\\ =\frac{12\mu \text{C}}{2\times 60.0\text{cm}}\\ =0.1\mu \text{C}/\text{cm}\end{array}$

Substitute $8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $k$, $0.1\mu \text{C}/\text{cm}$ for ${\lambda }_{\text{ο}}$, $60.0\text{cm}$ for $R$ in equation (VIII) to find $\overrightarrow{E}$.

    $\begin{array}{c}\overrightarrow{E}=-\frac{8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\times \pi \times \left(0.1\mu \text{C}/\text{cm}\right)\left(\frac{{10}^{-6}\text{C}}{1\mu \text{C}}\right)\left(\frac{100\text{cm}}{1\text{m}}\right)}{2\left(60.0\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)}\widehat{j}\\ =-\frac{8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\times \pi \times 10\times {10}^{-6}\text{C}/\text{m}}{2\left(0.6\text{m}\right)}\widehat{j}\\ =-235.36\times {10}^3\widehat{j}\text{N}/\text{C}\end{array}$

Substitute $-235.36\times {10}^3\widehat{j}\text{N}/\text{C}$ for $\overrightarrow{E}$ and $3.00\mu \text{C}$ for $q$ in the equation (I) to find $F$.

    $\begin{array}{c}F=3.00\mu \text{C}\left(-235.36\times {10}^3\widehat{j}\text{N}/\text{C}\right)\\ =\left(3.00\mu \text{C}\right)\left(\frac{{10}^{-6}\text{C}}{1\mu \text{C}}\right)\times \left(-235.36\times {10}^3\widehat{j}\text{N}/\text{C}\right)\\ =-0.706\widehat{j}\text{N}\end{array}$

Therefore, the total force on a charge of $3.00\mu \text{C}$ at the center of curvature is $-0.706\widehat{j}\text{N}$.