Question

derive an expression for electric field at a point on axial equitorial position of electric dipole

Answer 1

Consider an electric dipole consisting of two point changes −q−q and +q+q separated by a small distance AB=2aAB=2a with center O and dipole moment P=q(2a)P=q(2a)

Let OP=rOP=r

E1E1= electric intensity at P due to change -q at A.

∴|E1−→|=14π∈0qAP2∴|E1→|=14π∈0qAP2

AP2=OP2+OA2AP2=OP2+OA2

=r2+a2=r2+a2

∴|E1−→|=14π∈0qr2+a2∴|E1→|=14π∈0qr2+a2

E1−→=Pc−→E1→=Pc→

Let ∠PBA=∠PAB=θ∠PBA=∠PAB=θ.

E1E1 has two components E1E1 can θθ along PR∥BAPR∥BAand E1sinθE1sin⁡θ along PE⊥BAPE⊥BA .

If E2E2 is the electric intensity at P, due to change +q at B, then

E2−→=14π∈0qBp2=14π∈0q(r2+a2)E2→=14π∈0qBp2=14π∈0q(r2+a2)

E2−→E2→ is along PD−→−PD→

This has two components, E2E2 is along PR∥BAPR∥BAand E2sinθE2sin⁡θ along PF.

E→=2E1cosθE→=2E1cos⁡θ

=24π∈0.q(r2+q2)=24π∈0.q(r2+q2)cosθcos⁡θ

=24π∈0.q(r2+a2)(OAAD)=24π∈0.q(r2+a2)(OAAD)

=24π∈0.q(r2+q2)ar2+a2−−−−−√=24π∈0.q(r2+q2)ar2+a2

=q×2a4π∈0(r2+a2)3/2=q×2a4π∈0(r2+a2)3/2

But q×2a=|P→|q×2a=|P→|, the dipole moment

∴|E→|=|P→|4π∈0(r2+a2)3/2∴|E→|=|P→|4π∈0(r2+a2)3/2

The direction of E→E→ is along .

PR−→−||BA−→−PR→||BA→ (ie) opposite to P→P→

or E→=−P→4π∈0(r2+a2)3/2E→=−P→4π∈0(r2+a2)3/2

If the dipole is short 2a<<r2a<<r

|E→|=|P→|4π∈0r3|E→|=|P→|4π∈0r3

|E→|α1r3