गाउस नियम का उपयोग किए बिना किसी एकसमान रैखिक आवेश घनत्व $\lambda$ के लंबे पतले तार के कारण विध्युत क्षेत्र के लिए सूत्र प्राप्त कीजिए

Take a long thin wire $XY$ (as shown in the following figure) of uniform linear charge density $\lambda$

Consider a point $A$ at a perpendicular distance $l$ from the mid-point $O$ of the wire, as shown in the following figure.

Let $E$ be the electric field at point $A$ due to the wire,$ XY$.

Consider a small length element $d x$ on the wire section with $OZ =x$

Let $q$ be the charge on this piece.

$=\lambda d x$

Electric field due to the piece,

$d E=\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{\lambda d x}{(A Z)^{2}}$

However, $A Z=\sqrt{l^{2}+x^{2}}$

$\therefore d E=\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{\lambda d x}{\left(l^{2}+x^{2}\right)}$

The electric field is resolved into two rectangular components. $d E \cos \theta$ is the perpendicular component and $d E \sin \theta$ is the parallel component. When the whole wire is considered, the component $d E \sin \theta$ is cancelled. Only the perpendicular component $d E \cos \theta$ affects point $A$ Hence, effective electric field at point $A$ due to the element dx is $dE_{1}.$

In $\Delta AZO$

$\therefore d E_{1}=\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{\lambda d x \cdot \cos \theta}{\left(l^{2}+x^{2}\right)} \dots \ldots(1)$

$\tan \theta=\frac{x}{l} \Rightarrow x=l \cdot \tan \theta \ldots \ldots (2)$

On differentiating equation $(2),$ we obtain $\frac{d x}{d \theta}=l \sec ^{2} \theta \Rightarrow d x$$=l \sec ^{2} \theta d \theta \ldots \ldots (3)$

From equation $(2),$ we have $x^{2}+l^{2}=l^{2} \tan ^{2} \theta+l^{2}=l^{2}\left(\tan ^{2} \theta+1\right)$$=l^{2} \sec ^{2} \theta \ldots \ldots (4)$

Putting equations $(3)$ and $(4)$ in equation $(1),$ we obtain

$d E_{1}=\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{\lambda l \sec ^{2} \theta d \theta \cdot \cos \theta}{l^{2} \sec ^{2} \theta}$

$= \frac{1}{4 \pi \epsilon_{0}} \cdot \frac{\lambda \cos \theta d \theta}{l} \ldots \ldots(5)$

The wire is so long that $\theta$ tends from $-\frac{\pi}{2}$ to $\frac{\pi}{2} .$

By integrating equation $(5),$ we obtain the value of field $E _{1}$ as,

$\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} d E_{1}=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{4 \pi \epsilon_{0}} \frac{\lambda}{l} \cos \theta d \theta$

$\Rightarrow E_{1}=\frac{1}{4 \pi \epsilon_{0}} \frac{\lambda}{l}[\sin \theta]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}$

$\Rightarrow E_{1}=\frac{1}{4 \pi \epsilon_{0}} \frac{\lambda}{l} \times 2$

$\Rightarrow E_{1}=\frac{\lambda}{2 \pi \epsilon_{0} l}$

Therefore, the electric field due to long wire is $\frac{\lambda}{2 \pi \epsilon_{0} l}$