Chapter 24, Problem 79PQ

(a)

To determine

The expression for the electric field at point A located at a distance l above the mid-point of the rod.

Answer to Problem 79PQ

The expression for the electric field at point A located at a distance l above the mid-point of the rod is ${\overrightarrow{E}}_{tot}=\frac{\left(0.628\right)kQ}{l^2}\widehat{j}$.

Explanation of Solution

Sketch the diagram showing the five charges.

The x component of the electric field is zero based on the geometry.

Write the expression for the y component of the electric field.

    $\begin{array}{c}\overrightarrow{E}=\frac{kq}{r^2}\widehat{r}\\ =\frac{kq}{r^2}\sin \theta \widehat{j}\end{array}$                                                                                                  (I)

Here, $\overrightarrow{E}$ is the electric field, $k$ is the coulomb constant, $q$ is the charge and $r$ is the distance between the charge and the point.

Write the equation for the total electric field.

    ${\overrightarrow{E}}_{tot}={\overrightarrow{E}}_1+{\overrightarrow{E}}_2+{\overrightarrow{E}}_3+{\overrightarrow{E}}_4+{\overrightarrow{E}}_5$                                                                       (II)

Conclusion:

Substitute $\frac{Q}{5}$ for $q$, $\sqrt{l^2+l^2}$ for $r$ and $\frac{l}{\sqrt{l^2+l^2}}$ for $\sin \theta$ in equation (I) to find ${\overrightarrow{E}}_1$.

    $\begin{array}{c}{\overrightarrow{E}}_1=\frac{k\left(\frac{Q}{5}\right)}{{\left(\sqrt{l^2+l^2}\right)}^2}\frac{l}{\sqrt{l^2+l^2}}\widehat{j}\\ =\frac{kQ}{10\sqrt{2}{l}^2}\widehat{j}\end{array}$                                                                         (III)

Substitute $\frac{Q}{5}$ for $q$, $\sqrt{l^2+l^2}$ for $r$ and $\frac{l}{\sqrt{l^2+l^2}}$ for $\sin \theta$ in equation (I) to find ${\overrightarrow{E}}_2$.

    $\begin{array}{c}{\overrightarrow{E}}_2=\frac{k\left(\frac{Q}{5}\right)}{{\left(\sqrt{l^2+{\left(\frac{l}{2}\right)}^2}\right)}^2}\frac{l}{\sqrt{l^2+{\left(\frac{l}{2}\right)}^2}}\widehat{j}\\ =\frac{k\left(\frac{Q}{5}\right)}{\left(\frac{5}{4}\right)l^2}\frac{2}{\sqrt{5}}\widehat{j}\\ =\frac{8kQ}{25\sqrt{5}{l}^2}\widehat{j}\end{array}$                                                                (IV)

Substitute $\frac{Q}{5}$ for $q$ and $l$ for $r$ in equation (I) to find ${\overrightarrow{E}}_3$.

    $\begin{array}{c}{\overrightarrow{E}}_3=\frac{k\left(\frac{Q}{5}\right)}{l^2}\widehat{j}\\ =\frac{kQ}{5l^2}\widehat{j}\end{array}$                                                                                               (V)

Substitute $\frac{Q}{5}$ for $q$, $\sqrt{l^2+l^2}$ for $r$ and $\frac{l}{\sqrt{l^2+l^2}}$ for $\sin \theta$ in equation (I) to find ${\overrightarrow{E}}_4$.

    $\begin{array}{c}{\overrightarrow{E}}_4=\frac{k\left(\frac{Q}{5}\right)}{{\left(\sqrt{l^2+{\left(\frac{l}{2}\right)}^2}\right)}^2}\frac{l}{\sqrt{l^2+{\left(\frac{l}{2}\right)}^2}}\widehat{j}\\ =\frac{k\left(\frac{Q}{5}\right)}{\left(\frac{5}{4}\right)l^2}\frac{2}{\sqrt{5}}\widehat{j}\\ =\frac{8kQ}{25\sqrt{5}{l}^2}\widehat{j}\end{array}$                                                                (VI)

Substitute $\frac{Q}{5}$ for $q$, $\sqrt{l^2+l^2}$ for $r$ and $\frac{l}{\sqrt{l^2+l^2}}$ for $\sin \theta$ in equation (I) to find ${\overrightarrow{E}}_5$.

    $\begin{array}{c}{\overrightarrow{E}}_5=\frac{k\left(\frac{Q}{5}\right)}{{\left(\sqrt{l^2+l^2}\right)}^2}\frac{l}{\sqrt{l^2+l^2}}\widehat{j}\\ =\frac{kQ}{10\sqrt{2}{l}^2}\widehat{j}\end{array}$                                                                         (VII)

Substitute equations (III), (IV), (V), (VI) and (VII) in equation (II) to find ${\overrightarrow{E}}_{tot}$.

    $\begin{array}{c}{\overrightarrow{E}}_{tot}=\frac{kQ}{10\sqrt{2}{l}^2}\widehat{j}+\frac{8kQ}{25\sqrt{5}{l}^2}\widehat{j}+\frac{kQ}{5l^2}\widehat{j}+\frac{8kQ}{25\sqrt{5}{l}^2}\widehat{j}+\frac{kQ}{10\sqrt{2}{l}^2}\widehat{j}\\ =\frac{\sqrt{2}kQ}{10l^2}\widehat{j}+\frac{16kQ}{25\sqrt{5}{l}^2}\widehat{j}+\frac{kQ}{5l^2}\\ =\frac{kQ}{l^2}\left(\frac{\sqrt{2}}{10}+\frac{16}{25\sqrt{5}}+\frac{1}{5}\right)\\ =\frac{\left(0.628\right)kQ}{l^2}\widehat{j}\end{array}$

Thus, the expression for the electric field at point A located at a distance l above the mid-point of the rod is ${\overrightarrow{E}}_{tot}=\frac{\left(0.628\right)kQ}{l^2}\widehat{j}$.

(b)

To determine

The electric field at point A located at a distance l above the mid-point of the rod using the exact expression.

Answer to Problem 79PQ

The electric field at point A located at a distance l above the mid-point of the rod using the exact expression is ${\overrightarrow{E}}_{tot}=\frac{kQ}{\sqrt{2}{l}^2}\widehat{j}$.

Explanation of Solution

Write the exact expression for the total electric field.

    $\overrightarrow{E}=\frac{kq}{y}\frac{1}{\left(\sqrt{l^2+y^2}\right)}\widehat{j}$                                                                                    (VIII)

Here, $\overrightarrow{E}$ is the electric field, $k$ is the coulomb constant, $q$ is the charge, $y$ is the perpendicular distance between the rod and the point and $l$ is the length of the rod.

Conclusion:

Substitute $Q$ for $q$, and $l$ for $y$ in equation (VIII) to find ${\overrightarrow{E}}_{tot}$.

    $\begin{array}{c}{\overrightarrow{E}}_{tot}=\frac{kQ}{l}\frac{1}{\left(\sqrt{l^2+l^2}\right)}\widehat{j}\\ =\frac{kQ}{\sqrt{2}{l}^2}\widehat{j}\end{array}$

Thus, the electric field at point A located at a distance l above the mid-point of the rod using the exact expression is ${\overrightarrow{E}}_{tot}=\frac{kQ}{\sqrt{2}{l}^2}\widehat{j}$.

(c)

To determine

Compare the approximate result with the exact result.

Answer to Problem 79PQ

The approximate result is $0.888$ times the exact result.

Explanation of Solution

Find the ratio of the approximate result with the exact result.

    $\begin{array}{c}\frac{{\overrightarrow{E}}_{tot,a}}{{\overrightarrow{E}}_{tot,b}}=\frac{\frac{\left(0.628\right)kQ}{l^2}\widehat{j}}{\frac{kQ}{\sqrt{2}{l}^2}\widehat{j}}\\ =\left(0.628\right)\left(\sqrt{2}\right)\\ =0.888\end{array}$

Conclusion:

Thus, the approximate result is $0.888$ times the exact result.