Question

derive an expression for tje electtric field intensity at a point at distance r from a point electric charge?

Answer 1

Consider an electric dipole consisting of two point changes −q−q and +q+q separated by a small distance AB=2aAB=2awith center O and dipole moment P=q(2a)P=q(2a)Let OP=rOP=rE1E1= electric intensity at P due to change -q at A.∴|E1−→|=14π∈0qAP2∴|E1→|=14π∈0qAP2AP2=OP2+OA2AP2=OP2+OA2=r2+a2=r2+a2∴|E1−→|=14π∈0qr2+a2∴|E1→|=14π∈0qr2+a2E1−→=Pc−→E1→=Pc→Let ∠PBA=∠PAB=θ∠PBA=∠PAB=θ.E1E1 has two components E1E1 can θθ along PR∥BAPR∥BA and E1sinθE1sin⁡θ along PE⊥BAPE⊥BA .If E2E2 is the electric intensity at P, due to change +q at B, thenE2−→=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∥BA and 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)3/2But q×2a=|P→|q×2a=|P→|, the dipole moment∴|E→|=|P→|4π∈0(r2+a2)3/2∴|E→|=|P→|4π∈0(r2+a2)3/2The 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)3/2If the dipole is short 2a<<r2a<<r|E→|=|P→|4π∈0r3|E→|=|P→|4π∈0r3|E→|α1r3

Answer 2

in this way we find " electric field intensity due to point charge".