2.3 Find the electric field a distance \(z\) above one end of a straight line segment of length \(L\), which carries a uniform line charge \(\lambda\). Check that your formula is consistent with what you would expect for the case \( z \gg L \).
$$d \vec{E} = \frac{ 1 }{4 \pi \epsilon_0} \frac{dq}{z^2 + l^2} \cdot \vec{u} = \frac{ 1 }{4 \pi \epsilon_0} \frac{\lambda dl}{z^2 + l^2} \left( \frac{-l}{\sqrt{z^2+l^2}} \hat{i} + \frac{z}{\sqrt{z^2+l^2}} \hat{j} \right)$$

$$ \vec{E} = \frac{ \lambda }{4 \pi \epsilon_0} \int_0^L \frac{-l}{\left(z^2+l^2\right)^{3/2}} \, dl \, \hat{i} + \frac{ \lambda }{4 \pi \epsilon_0} \int_0^L \frac{z}{\left(z^2+l^2\right)^{3/2}} \, dl \, \hat{j}$$

The integration of the i-hat component is a straight substitution; however, the j-hat component requires a trigonometric substitution and will review that technique below.

$$\frac{ \lambda }{4 \pi \epsilon_0} \int_0^L \frac{z}{\left(z^2+l^2\right)^{3/2}} \, dl \, \hat{j}$$

$$l = z \tan \theta \hspace{10mm} dl = z \sec^2 \theta d\theta$$

$$\frac{ \lambda }{4 \pi \epsilon_0} \int_0^{\tan^{-1} (L/z)} \frac{z^2 \sec^2 \theta}{\left(z^2+z^2 \tan^2 \theta \right)^{3/2}} \, d\theta \, \hat{j}$$

$$\frac{ \lambda }{4 \pi \epsilon_0} \int_0^{\tan^{-1} (L/z)} \frac{z^2 \sec^2 \theta}{z^3 \left(1+ \tan^2 \theta \right)^{3/2}} \, d\theta \, \hat{j}$$

$$\frac{ \lambda }{4 \pi \epsilon_0} \int_0^{\tan^{-1} (L/z)} \frac{ \sec^2 \theta}{z \left(1+ \tan^2 \theta \right)^{3/2}} \, d\theta \, \hat{j}$$

$$\frac{ \lambda }{4 \pi \epsilon_0} \int_0^{\tan^{-1} (L/z)} \frac{1}{z \sec \theta} \, d\theta \, \hat{j}$$

$$\frac{ \lambda }{4 \pi \epsilon_0 z} \left( \sin \left( \tan^{-1} \frac{L}{z} \right) – 0 \right) \hat{j} = \frac{ \lambda }{4 \pi \epsilon_0 z} \frac{L}{z \sqrt{L^2+z^2}} \, \hat{j} $$

Adding the components of the electric field yields

$$\vec{E} = \frac{ \lambda }{4 \pi \epsilon_0} \left[\left( \frac{-1}{\sqrt{z^2}}+\frac{1}{\sqrt{L^2+z^2}} \right) \hat{i}+ \frac{L}{z \sqrt{L^2+z^2}} \, \hat{j} \right] = \frac{ \lambda }{4 \pi \epsilon_0 z} \left[\left( -1+\frac{z}{\sqrt{L^2+z^2}} \right) \hat{i}+ \frac{L}{\sqrt{L^2+z^2}} \, \hat{j} \right]$$