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Question
An electric dipole of dipole moment`vecp` consists of point charges +q and −q separated by a distance 2a apart. Deduce the expression for the electric field `vecE` due to the dipole at a distance x from the centre of the dipole on its axial line in terms of the dipole moment `vecp`. Hence show that in the limit x>> a, `vecE->2vecp"/"(4piepsilon_0x^3)`
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Solution
Electric Field on Axial Line of an Electric Dipole
Let P be at distance r from the centre of the dipole on the side of charge −q. Then, the electric field at point P due to charge −q of the dipole is given by
`vecE_-q=-q/(4piepsilon_0(r+a)^2)hatp`
Where `hatp `is the unit vector along the dipole axis (from − q to q).
Also, the electric field at point P due to charge +q of the dipole is given by
`vecE_+q=q/(4piepsilon_0(r-a)^2)hatp`
The total field at P is
`vecE=vecE_+q+vecE_-q=q/(4piepsilon_0)[1/(r-a)^2-1/(r+a)^2]hatp`
`=>vecE=q/(4piepsilon_0)(4ar)/(r^2-a^2)^2hatp `
Given:
r = x
`vecE=q/(4piepsilon_0)(4ax)/(x^2-a^2)^2hatp`
For x >> a,
`vecE=(4qa)/(4piepsilon_0x^3)hatp`
`vecE=(2vecp)/(4piepsilon_0x^3)" "[.:vecp=(qxx2a)hatp]`
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