Question

Write the magnitude and direction of electric field intensity due to an electric dipole of length 2a and

Answer 1

vec E = - (2kq)/a^2 haty

Explanation:

E = kq/r^2

In the middle of the dipole, the electric fields contributed by both charges are in the same direction; they are both pointing toward the negative charge. Thus, the magnitude of the electric field is the sum of the contribution of the two charges.

E = kq/a^2 + kq/a^2 = 2kq/a^2

The direction of the field is pointing to the negative charge. So its actual direction depends on how the dipole is oriented. If the dipole is oriented north-south and that the +q is the northern pole, then E is pointing to the south. That is,

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 ,

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