Find the electric field a distance zabove one end of a straight line segment of length L(Fig. 2.7) that carries a uniform line charge A. Check that your formula is consistent with what you would expect for the case z»L.

A straight segment of length$L$ carries a uniform line charge of magnitude.The electric force exerted by line segment above one end at a distance z on the z-axis,has to be evaluated.

Electric force exerted by charge $\mathit{q}$ on charge $\mathbf{Q}$is proportional to the product of the charge $\mathit{q}$and inversely proportional to the square of the distance between them as,

$\mathbf{E}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\frac{\mathbf{q}}{{\mathbf{r}}^{\mathbf{2}}}$

A straight segment of length $L$carries a uniform line charge of magnitude$A$.is shown in following figure, along with the components of electric field at the point on the z axis.

Rod has its one end lying on the origin and is placed parallel to x axis. The differential element along the length of the rod be $dx$, such that the differential charge on the rod can be written as

$dq=\lambda dx$ ……. (1)

Here $\lambda$is the line charge density.

From the above figure, from the right triangle, using Pythagoras theorem we can write

$r=\sqrt{x^2+Z^2}$ …… (2)

Also, from the figure,

$\cos \theta =\frac{z}{\sqrt{X^2+Z^2}}$

$\sin \theta =\frac{x}{\sqrt{X^2+z^2}}$

Theelectric field at the point, along the z direction and x direction can be written as,

$d{E}_z=dE\cos \theta$

$d{E}_x=dE\sin \theta$

The electric field along the rod above its one end has both x and z component. So, the resultant differential electric filed can be written as

$$\begin{gathered} dE=-dE\sin \theta \hat{x}+dE\cos \theta \hat{z} \\ =-\left(\frac{1}{4\pi {\epsilon }_0}\frac{dq}{x^2+z^2}\right)\sin \theta \hat{x}+\left(\frac{1}{4\pi {\epsilon }_0}\frac{dq}{x^2+z^2}\right)\cos \theta \hat{z} \end{gathered}$$ …… (2)

Substitutelocalid="1655871895446" $\lambda dx\mathrm{for}dq,\frac{z}{\sqrt{x^2}+z^2}for\cos \theta \mathrm{and}\frac{x}{\sqrt{x^2}+z^2}\mathrm{for}\mathrm{sin\theta }$into equation (2).

$$\begin{gathered} dE=-\left(\frac{1}{4\pi {\epsilon }_0}\frac{\lambda dx}{x^2+z^2}\right)\left(\frac{x}{\sqrt{x^2+z^2}}\right)\hat{x}+\left(\frac{1}{4\pi {\epsilon }_0}\frac{\lambda dx}{x^2+z^2}\right)\left(\frac{z}{\sqrt{x^2+z^2}}\right)\hat{z} \\ =\left(\frac{1}{4\pi {\epsilon }_0}\frac{\lambda dx}{x^2+z^2}\right)\left(-\frac{xdx}{\sqrt{x^2+z^2}}\hat{x}+\frac{zdx}{\sqrt{x^2+z^2}}\hat{z}\right) \end{gathered}$$

Integrate the above equation to obtain the total electric field due to total length of the rod, above its one end, with the limit from 0 to localid="1654513712781" $L$.

$$\begin{gathered} E={\int }_0^L\frac{\lambda }{4\pi {\epsilon }_0}\left(\frac{xdx}{{\left(x^2+z^2\right)}^{3/2}}\hat{x}+\frac{zdx}{{\left(x^2+z^2\right)}^{3/2}}\hat{z}\right) \\ =\frac{\lambda }{4\pi {\epsilon }_0}\left\{-{\left(-\frac{1}{\sqrt{x^2+z^2}}\right)}_0^L\hat{x}+z{\left(\frac{x}{\sqrt{x^2+z^2}}\right)}_0^L\hat{z}\right\} \\ =\frac{\lambda }{4\pi {\epsilon }_0}\left\{{\left(\frac{1}{\sqrt{L^2+z^2}}\right)}_0^L\hat{x}+z{\left(\frac{x}{z^2\sqrt{L^2+z^2}}-0\right)}_0^L\hat{z}\right\} \\ =\frac{\lambda }{4\pi {\epsilon }_{0}z}\left[\left(\frac{z}{\sqrt{L^2+z^2}}-1\right)\hat{x}+\left(\frac{L}{\sqrt{L^2+z^2}}\right)\hat{z}\right] \end{gathered}$$

Thus the resultant electric field due to the rod above its one end is

$E=\frac{\lambda }{4\pi {\epsilon }_{0}z}\left[\left(\frac{z}{\sqrt{L^2+z^2}}-1\right)\hat{x}+\left(\frac{L}{\sqrt{L^2+z^2}}\right)\hat{z}\right]$

Determine the electric field for localid="1655873310206" $z\gg L$at the point P.

$$\begin{gathered} E=\frac{\lambda }{4\pi {\epsilon }_{0}z}{\left[1-\frac{z}{\sqrt{z^2\left(1+{\displaystyle \frac{L^2}{z^2}}\right)}}\right]}^{\frac{1}{2}} \\ =\frac{\lambda }{4\pi {\epsilon }_{0}z}\left\{1-{\left(1+\frac{L^2}{z^2}\right)}^{\frac{-1}{2}}\right\} \end{gathered}$$

Simply using binomial theorem as,

role="math" localid="1655873489047" $\begin{array}{l}E=\frac{\lambda L}{4\pi {\epsilon }_0{z}^2}\\ =\frac{q}{4\pi {\epsilon }_0{z}^2}\end{array}$

Therefore, the expression for the electric field is$\frac{q}{4\pi {\epsilon }_0{z}^2}$.