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प्रश्न
Show that intensity of electric field at a point in broadside position of an electric dipole is given by:
E = `(1/(4piepsilon_0)) "p"/(("r"^2 + l^2)^(3//2))`
Where the terms have their usual meaning.
उत्तर
Electric field due to dipole (broadside on position)
Consider dipole of length 2l and moment `vecp`.
From the figure, resultant electric field intensity at C:
`vec"E" = vec"E"_"A" + vec"E"_"B"`
`|vec"E"_"A"| = 1/(4piepsilon_0) "q"/("r"^2 + l^2)`
`|vec"E"_"B"| = 1/(4piepsilon_0) "q"/("r"^2 + l^2)`
Sine components of `vec"E"_"A" "and" vec"E"_"B"` get cancelled each other as `|vec"E"_"A"| = |vec"E"_"B"|`.
The cosine components get added up to give the resultant field.
i.e., E = EA cos θ + EB cos θ
`= 1/(4piepsilon_0) * "q"/("r"^2 + l^2) cos theta + 1/(4piepsilon_0) * "q"/("r"^2 + l^2) cos theta`
`= 2 * 1/(4piepsilon_0) * "q"/("r"^2 + l^2) cos theta`
`= 2 xx 1/(4piepsilon_0) * "q"/("r"^2 + l^2) xx l/("r"^2 + l^2)^(1//2)`
`= 1/(4piepsilon_0) ("q"(2l))/(("r"^2 + l^2)^(3//2))`
E = `1/(4piepsilon_0) "p"/(("r"^2 + l^2)^(3//2))` as p = q(2l)
The above expression gives the magnitude of the field. The direction of electric field E at C is opposite to the direction of the dipole moment `vec"p"`.
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