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Question
Answer the following question.
Derive an expression for the electric field due to a dipole of dipole moment `vec"p"` at a point on its perpendicular bisector.
Solution
The magnitudes of the electric field due to the two charges +q and −q are given by,
`E_(+q) = ("q")/(4πε_0) (1)/("r"^2 + a^2)` .....(i)
`E_(-q) = ("q")/(4πε_0) (1)/("r"^2 + a^2)` .....(i)
∴ `E_(+q) = E_(-q)`
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 hat"p" ... ["negative sign shows that field is opposite to" hat"p"]`
`E = -(2qa)/(4πε_0("r"^2 + a^2)^(3/2)) hat"p"` ....(iii)
At large distances (r >> a), this reduces to
`E = -(2qa)/(4πε_0"r"^3)hat"p"` ....(iv)
∵ `vec"p" = q xx vec"2a"hat"p"`
∴ `E = (-vec"p")/(4πε_0"r"^3)` ...(r>>a).
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