Obtain the formula for the electric field due to a long thin wire of uniform linear charge density λ without using Gauss’s law. [Hint: Use Coulomb’s law directly and evaluate the necessary integral.] 

Consider a long thin wire of uniform linear charge density, $\mathrm{\lambda }$. 

To find: Formula for electric field due to this wire at any point P at a perpendicular distance PC = r from the wire. 

Consider a small element of length dx of the wire with centre O, such that OC = x. 

Charge on the element, q = $\mathrm{\lambda }.\mathrm{dx}$ 

So, electric intensity at P due to the element is given by,  

$\mathrm{dE}= \frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\frac{\mathrm{\lambda }.\mathrm{dx}}{{\mathrm{OP}}^2}\mathrm{ }=\mathrm{ }\frac{\mathrm{\lambda }\mathrm{ }.\mathrm{ }\mathrm{dx}}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}({\mathrm{r}}^2\mathrm{ }+\mathrm{ }{\mathrm{x}}^2)}$ 

Now, $\stackrel{\rightharpoonup }{\mathrm{dE}}$ can be resolved into two rectangular components, that is $\stackrel{\rightharpoonup }{\mathrm{dE}} \cos \mathrm{\theta }$ in a perpendicular direction and $\stackrel{\rightharpoonup }{\mathrm{dE}} \sin \mathrm{\theta }$ in a parallel direction. 

The parallel component will be cancelled by the parallel component of the field due to charge on a similar element dx of wire on the other half.

The radial components get added. 

Therefore, 

Effective component of electric intensity due to the charge element, $\mathrm{dE}\text{'} =\stackrel{\rightharpoonup }{\mathrm{dE}}\cos \mathrm{\theta }$ 

$\mathrm{dE}\text{'} =\mathrm{ }\frac{\mathrm{\lambda }.\mathrm{dx}\mathrm{ }\cos \mathrm{ }\mathrm{\theta }}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}\mathrm{ }({\mathrm{r}}^2+{\mathrm{x}}^2)}$                         ...(1) 

From $∆\mathrm{ }\mathrm{OCP},\mathrm{ }\mathrm{x}\mathrm{ }=\mathrm{ }\mathrm{r}\mathrm{ }\tan \mathrm{ }\mathrm{\theta }$

$\therefore \mathrm{ }\mathrm{dx}\mathrm{ }=\mathrm{ }\mathrm{r}\mathrm{ }{\sec }^2\mathrm{\theta }\mathrm{ }\mathrm{d\theta }$ 

$\mathrm{Now},\mathrm{ }{\mathrm{r}}^2 +\mathrm{ }{\mathrm{x}}^2 =\mathrm{ }{\mathrm{r}}^2 +\mathrm{ }{\mathrm{r}}^2\mathrm{ }\tan { }^2\mathrm{ }\mathrm{\theta }\mathrm{ }=\mathrm{ }{\mathrm{r}}^2 (1+\mathrm{ }{\tan }^2 \mathrm{\theta })\mathrm{ }=\mathrm{ }{\mathrm{r}}^2 {\sec }^2 \mathrm{\theta }$

From equation (1), we have 

$\mathrm{dE}\text{'} =\mathrm{ }\frac{\mathrm{\lambda }\mathrm{ }\mathrm{r}\mathrm{ }{\sec }^2\mathrm{\theta }\mathrm{ }\mathrm{d\theta }}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}{\mathrm{r}}^2\mathrm{ }{\sec }^2\mathrm{\theta }}\cos \mathrm{ }\mathrm{\theta }$ 

$\Rightarrow \mathrm{ }\mathrm{dE}\text{'} =\mathrm{ }\frac{\mathrm{\lambda }}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}\mathrm{ }\mathrm{r}\mathrm{ }}\mathrm{ }\cos \mathrm{ }\mathrm{\theta }\mathrm{ }\mathrm{d\theta }$ 

Since the wire has infinite length, it’s ends A and B are infinite distances apart. 

Therefore, $\mathrm{\theta }$ varies from $-\frac{\mathrm{\pi }}{2}\mathrm{to} +\frac{\mathrm{\pi }}{2}$

So, Electric Intensity at P due to the whole wire is given by, 

      $\mathrm{E}\text{'} ={\int }_{-\frac{\mathrm{\pi }}{2}}^{+\frac{\mathrm{\pi }}{2}}\frac{\mathrm{\lambda }}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}\mathrm{ }\mathrm{r}\mathrm{ }}\cos \mathrm{ }\mathrm{\theta }\mathrm{ }\mathrm{d\theta }$ 

$\Rightarrow \mathrm{ }\mathrm{E}\text{'} =\mathrm{ }\frac{\mathrm{\lambda }}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}\mathrm{ }\mathrm{r}\mathrm{ }}{\left[\sin \mathrm{ }\mathrm{\theta }\right]}_{-\frac{\mathrm{\pi }}{2}}^{+\frac{\mathrm{\pi }}{2}}$ 

         $=\mathrm{ }\frac{\mathrm{\lambda }}{2{\mathrm{\pi \epsilon }}_{\mathrm{o}}\mathrm{ }\mathrm{r}\mathrm{ }}$, is the required electric field intensity.