Chapter 16, Problem 22E

(a)

To determine

To show that the electric field normal to the wire of length extending from $-L$ to $+L$ is $E=\frac{2k\lambda L}{r{\left(r^2+L^2\right)}^{1/2}}$.

Answer to Problem 22E

The electric field normal to the wire of length extending from $-L$ to $+L$ is $E=\frac{2k\lambda L}{r{\left(r^2+L^2\right)}^{1/2}}$.

Explanation of Solution

Observe the Figure 16.18.

Write the expression for the linear charge density of the wire.

  $\lambda =\frac{Q}{L}$        (I)

Here, $\lambda$ is the linear charge density, $Q$ is the charge, and $L$ is the length of wire.

For an elemental length $dx$, the charge is $\lambda dx$.

From Figure 16.18a, the distance to an arbitrary point $P$ is $R$, where $R$ is given by,

  $R={\left(r^2+x^2\right)}^{1/2}$        (II)

This gives the magnitude of electric field due to the charge $\lambda dx$ as,

  $dE=\frac{k\lambda dx}{R^2}$        (III)

Use equation (II) in (III).

  $dE=\frac{k\lambda dx}{\left(r^2+x^2\right)}$        (IV)

The total field can be resolved into horizontal and vertical components given by,

  $d{E}_x=dE\cos \theta$

  $d{E}_y=dE\sin \theta$

Among them the horizontal components the electric field due to two elementary lengths of the wire equidistant from the center cancel each other. Thus, $E_x$ is zero and the net electric field is contributed by only the vertical component of the field.

From Figure 16.18b, $\sin \theta$ can be deduced as,

  $\sin \theta =\frac{r}{{\left(r^2+x^2\right)}^{1/2}}$

The vertical component of the elementary field is thus obtained as,

  $d{E}_y=\frac{kr\lambda dx}{{\left(r^2+x^2\right)}^{3/2}}$        (V)

Thus the net electric field is obtained by integrating over the length of the wire. In this case the wire extends from $-L$ to $+L$. Thus integrate equation (V) in this limit to find the net electric field.

  $\begin{array}{c}E={\displaystyle \underset{-L}{\overset{+L}{\int }}d{E}_y}\\ ={\displaystyle \underset{-L}{\overset{+L}{\int }}\frac{kr\lambda dx}{{\left(r^2+x^2\right)}^{3/2}}}\\ =k\lambda {\displaystyle \underset{-L}{\overset{+L}{\int }}\frac{r}{{\left(r^2+x^2\right)}^{3/2}}dx}\\ =\frac{k\lambda }{r}{\displaystyle \underset{-L}{\overset{+L}{\int }}\frac{r^2}{{\left(r^2+x^2\right)}^{3/2}}dx}\end{array}$        (VI)

Use the standard integration formula Equation B.46 in Appendix B of the text book.

  ${\displaystyle \int \frac{a^2}{\left(a^2+t^2\right)}}dt=\frac{t}{{\left(a^2+t^2\right)}^{1/2}}$        (VII)

Use equation (VII) for determining the integral in equation (VI).

  $\begin{array}{c}E=\frac{k\lambda }{r}{\left[\frac{x}{{\left(r^2+x^2\right)}^{1/2}}\right]}_{-L}^{+L}\\ =\frac{k\lambda }{r}\left(\frac{L}{{\left(r^2+L^2\right)}^{1/2}}-\frac{\left(-L\right)}{{\left(r^2+{\left(-L\right)}^2\right)}^{1/2}}\right)\\ =\frac{k\lambda }{r}\frac{2L}{{\left(r^2+L^2\right)}^{1/2}}\\ =\frac{2k\lambda L}{r{\left(r^2+L^2\right)}^{1/2}}\end{array}$        (VIII)

This shows that the electric field normal to the wire of length extending from $-L$ to $+L$ is $E=\frac{2k\lambda L}{r{\left(r^2+L^2\right)}^{1/2}}$.

Conclusion:

Therefore, the electric field normal to the wire of length extending from $-L$ to $+L$ is $E=\frac{2k\lambda L}{r{\left(r^2+L^2\right)}^{1/2}}$.

(b)

To determine

To show that the equation $E=\frac{2k\lambda L}{r{\left(r^2+L^2\right)}^{1/2}}$ reduces to $E=\frac{2k\lambda }{r}$ for the infinitely long wire.

Answer to Problem 22E

The equation $E=\frac{2k\lambda L}{r{\left(r^2+L^2\right)}^{1/2}}$ reduces to $E=\frac{2k\lambda }{r}$ for the infinitely long wire.

Explanation of Solution

For infinitely long wire the integration limits in equation (VI) becomes $-\infty$ to $+\infty$. Use equation (VII) to perform the integration.

  $\begin{array}{c}E=\frac{k\lambda }{r}{\left[\frac{x}{{\left(r^2+x^2\right)}^{1/2}}\right]}_{-\infty }^{+\infty }\\ =\frac{k\lambda }{r}\left[1-\left(-1\right)\right]\\ =\frac{2k\lambda }{r}\end{array}$

Conclusion:

Therefore, the equation $E=\frac{2k\lambda L}{r{\left(r^2+L^2\right)}^{1/2}}$ reduces to $E=\frac{2k\lambda }{r}$ for the infinitely long wire.