Question

Electric field intensity due to infinite line charge derivation

Answer 1

Hey mate here is your answer》》

The electric field of an infinite line charge with a uniform linear charge density can be obtained by a using Gauss' law. Considering a Gaussian surface in the form of a cylinder at radius r, the electric field has the same magnitude at every point of the cylinder and is directed outward.

Hope this answer will help you..《《

Answer 2

Electric Field Intensity :

The electric field intensity of any point is defined as the force experienced by unit positive charge placed at that point.

$\large\tt\red{E=\frac{F}{q_0}}$

The SI unit of electric field intensity

$\large \tt\orange{NC^{-1}}$

Electric field at any equatorial point of a dipole

$\large \tt{}is \: \: \: \green {\vec{E}=\frac{1}{4\pi\:\varepsilon_0}\frac{\vec{p}}{(r^2+a^2)^\frac{3}{2}}}$

At the mid-point of the dipole, r=0 so ,

$\large \tt \purple{\vec{E}=\frac{1}{4\pi\:\varepsilon_0}\frac{\vec{p}}{a^3}}$