Question

Derive an expression for the electric field intensity at a point on the equatorial line of an electric dipole moment →P and length 2a. What is the direction of this field ?

Answer 1

Hi...


Consider an electric dipole consisting of two point changes −q and +q separated by a small distance AB=2a with center O and dipole moment P=q(2a)
Let OP=r
E1= electric intensity at P due to change -q at A.
∴|E1−→|=14π∈0qAP2
AP2=OP2+OA2
=r2+a2
∴|E1−→|=14π∈0qr2+a2
E1−→=Pc−→
Let ∠PBA=∠PAB=θ.
E1 has two components E1 can θ along PR∥BA and E1sinθ along PE⊥BA .
If E2 is the electric intensity at P, due to change +q at B, then
E2−→=14π∈0qBp2=14π∈0q(r2+a2)
E2−→ is along PD−→−
This has two components, E2 is along PR∥BA and E2sinθ along PF.
E→=2E1cosθ
=24π∈0.q(r2+q2)cosθ
=24π∈0.q(r2+a2)(OAAD)
=24π∈0.q(r2+q2)ar2+a2−−−−−√
=q×2a4π∈0(r2+a2)3/2
But q×2a=|P→|, the dipole moment
∴|E→|=|P→|4π∈0(r2+a2)3/2
The direction of E→ is along .
PR−→−||BA−→− (ie) opposite to P→
or E→=−P→4π∈0(r2+a2)3/2
If the dipole is short 2a<|E→|=|P→|4π∈0r3
|E→|α1r3...

Hope this helps u!!

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 ,

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