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Obtain the formula for the electric field due to a long thin wire of uniform linear charge density $\lambda$ without using Gauss's law. [Hint: Use Coulomb's law directly and evaluate the necessary integral.]
PhysicsElectrostatics
Solution:
2258 Upvotes Verified Answer
Let $\mathrm{AB}$ be a thin wire having charge density (linear) $\lambda$. $\mathrm{D}$ is any point at normal distance $\mathrm{r}$ from a point $\mathrm{C}$, in the middle of the wire.


Let ' $\mathrm{dx}$ ' be a small element of wire $\mathrm{AB}$ \& at a distance $\sqrt{\mathrm{r}^2+\mathrm{x}^2}$ from point D.
Now, electric intensity at D due to charge on the element $\mathrm{dx}$ only
$\mathrm{E}_{\mathrm{DF}}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\lambda \mathrm{dx}}{\mathrm{r}^2+\mathrm{x}^2}$ along DF
Component along DG,
$\mathrm{E}_{\mathrm{DG}}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\lambda \mathrm{dx}}{\mathrm{r}^2+\mathrm{x}^2} \cos \theta$ and
$\mathrm{E}_{\mathrm{DH}}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\lambda \mathrm{dx}}{\left(\mathrm{r}^2+\mathrm{x}^2\right)} \sin \theta$
The parallel component $\mathrm{E}_{\mathrm{DH}}$ will be cancelled by the parallel component of the field of the charge by a similar element $\mathrm{dx}$ on the other half. The radial components gets added.

Radial component of electric intensity due to charge on element dx.
$$
\mathrm{dE}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\lambda \mathrm{dx}}{\left(\mathrm{r}^2+\mathrm{x}^2\right)} \cos \theta
$$
From the fig.
$$
\begin{aligned}
& x=r \tan \theta \\
& d x=r \sec ^2 \theta d \theta \\
\Rightarrow &\left(r^2+x^2\right)=\tan ^2 \theta=r^2\left(1+\tan ^2 \theta\right) \\
&\left(r^2+x^2\right)=r^2 \sec ^2 \theta
\end{aligned}
$$
Putting equation (ii) \& (iii) in equation (i), we have $\mathrm{dE}=\frac{\lambda}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{rsec}^2 \theta \cdot \mathrm{d} \theta}{\mathrm{r}^2 \sec ^2 \theta} \cos \theta=\frac{\lambda}{4 \pi \varepsilon_0 \mathrm{r}} \cos \theta \mathrm{d} \theta$
Since, wire has infinite length, its end $\mathrm{A}$ and $\mathrm{B}$ are quit away.
Hence ' $\theta$ ' varies from $-\pi / 2$ to $+\pi / 2$
$$
\begin{aligned}
&\Rightarrow \int_{-\pi / 2}^{+\pi / 2} \frac{\lambda}{4 \pi \varepsilon_0 \mathrm{r}} \cos \theta d \theta \\
&=\frac{\lambda}{4 \pi \varepsilon_0 \mathrm{r}}[\sin \theta]_{-\pi / 2}^{\pi / 2} \Rightarrow E=\frac{\lambda}{2 \pi \varepsilon_0 \mathrm{r}}
\end{aligned}
$$

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