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Derive an Expression for the Electric Field Due to a Dipole of Dipole Moment → P at a Point on Its Perpendicular Bisector. - Physics

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Answer in Brief

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.

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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).

Concept: Electric Field Lines
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