Question

Obtain the formula for electric field due to a long thin wire of uniform linear charge density without using gauss law.

Answer 1

Let a infinite long thin wire of surface charge density $\lambda$ is placed vertically as shown in figure. consider a point P, a unit away from the long charged wire. electric field due to element dy ,
$dE_p=\frac{Q}{4\pi\epsilon_0r^2}$ 
we know, charge = lines charge density × length 
So, $Q=\lambda dy$
r² = y² + a² [ from Pythagoras theorem, ]
$\implies dE_P=\frac{\lambda dy}{4\pi\epsilon_0(y^2 + a^2)}$
$E_p=\int\limits^{\infty}_{-\infty}{dE_p}$

here you can see that vertical components cancel out and horizontal component are added due to symmetry. 
$\implies E_P=\int\limits^{\infty}_{0}{2dE_pcos\theta}$

$\implies E_p=\int\limits^{\infty}_{0}{\frac{\lambda dy}{4\pi\epsilon_0(y^2+a^2)}cos\theta}$

$=\frac{2}{4\pi\epsilon_0}\int\limits^{\infty}_0{\frac{\lambda dy}{y^2+a^2}}\times\frac{y}{\sqrt{y^2+a^2}}$ 

$=\frac{1}{2\pi\epsilon_0}\int\limits^{\infty}_0{\frac{\lambda ydy}{(y^2+a^2)^{3/2}}}$

take y² + a² = x 
differentiate with respect to x
2ydy = dx 
and taking proper limits
I mean, take upper limit y = ∞² + a² = ∞
and lower limit , y = 0² + a² = a²

$E_p=\frac{\lambda}{4\pi\epsilon_0}\int\limits^{\infty}_{a^2}{\frac{dx}{x^{3/2}}}$

$\implies\left[\begin{array}{c}\frac{-2\lambda}{4\pi\epsilon_0}\frac{1}{\sqrt{x}}\end{array}\right]^{\infty}_{a^2}= \frac{\lambda}{2\pi\epsilon_0 a}$

hence, $E=\frac{\lambda}{2\pi\epsilon_0 a}$

Answer 2

Answer:

The simplest way to do this derivation without hactic differentiation

Explanation:

May it help you