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Question
Drive the expression for electric field at a point on the equatorial line of an electric dipole.
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Solution
Electric Field for Points on the Equatorial Plane:
The magnitudes of the electric field due to the two charges +q and −q are given by,
`E_(+q)=q/(4piepsilon_0)1/(r^2+a^2)` .....(i)
`E_(-q)=q/(4piepsilon_0)1/(r^2-a^2)` .....(ii)
The directions of E+q and E−q are as shown in the figure. The components normal to the dipole axis cancel away. The components along the dipole axis add up.
∴ Total electric field
`E=-(E_(+q)+E_(-q))cos theta.hatp`[Negative sign shows that field is opposite to `hatp`]
`E=-(2qa)/(4piepsilon_0(r^2+a^2)^(3/2))hatp` .....(iii)
At large distances (r >> a), this reduces to
`E=-(2qa)/(4piepsilon_0(r^3))hatp` .....(iv)
`because vecp=qxxvec(2a)hatp`
`therefore E=(-vecp)/(4piepsilon_0(r^3))` (r >> a)
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