Electric field intensity due to electric dipole in its end-on position

*“Electric field intensity due to electric dipole in its end-on position”:

*Note: Please replace (a) into (d) in above figure. Thank u

Considered AB is electric dipole which consisting by two point charges (-q)  and (+q)  and of length (2d). This the dipole moment will be p = 2dq. We are to find the electric field intensity at point P at a distance  r  from the midpoint of dipole in axial position . let the dipole is placed in a medium of dielectric constant be  k.

The electric fiedl at the point  p due to +q is,

                              E1​=4πε0​1​(r−d)2q​ (along BP)

And  The electric field at the point p due to  -q  is,

                              E2​=4πε0​1​(r+d)2q​ (along PA)

It is clear from above expression that E1>E2 but E1 and E2 are in opposite direction thus the resultant intensity at point p will be —

                                 E=E1​+(−E2​)

E=[4πε0​1​(r−d)2q​−4πε0​1​(r+d)2q​] along BP

E=4πε0​q​[(r−d)21​−(r+d)21​] along BP

E=4πε0​q​[(r2−d2)24rd​] along BP
0​

​





​According to the definition of dipole —

d<<r  then  d²≈0

∴E=4πε0​q​−r44rd​=4πε0​q​r34d​

E=4πε0​1​r32p​   (along BP)  NC–¹

  [∵ Electric dipole moment p=q×2d]