Obtain the formula for the electric field due to a long thin wire of uniform linear charge density $E$ without using Gauss’s law.
Obtain the formula for the electric field due to a long thin wire of uniform linear charge density $E$ without using Gauss’s law.
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 $\frac{\lambda}{2 \pi \epsilon_{0} l}$
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