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Derive an expression for intensity of el...

Derive an expression for intensity of electric field are a point in broadside position or on an equatorial line of an electric dipole.

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Suppose that the point P is situated on the right bisector of the dipole AB at a distance r metre from its mid-point [Fig. (a)].
Let `E_(1) and E_(2)` be the magnitudes of the intensities of the electric field at P due to the charges + q and - 2 of the dipole respectively. The distance of P from each charge is `sqrt(r^(2)+l^(2))`. Therefore,
`E_(1)=(1)/(4pi epsi_(0)) (q)/((r^(2)+l^(2)))` away from +q
and `E_(2)=(1)/(4pi epsi_(0))(q)/((r^(2)+l^(2)))`. towards -q
The magnitudes of `E_(1) and E_(2)` are equal t(but direction are different). On resolving `E_(1)and E_(2)` into two components perpendicular to AB `(E_(1) sin theta and E_(2) sin theta)` cancel other (beucse they are equal and opposite), while the components par4allel to AB `(E_(1) cos theta E_(2) cos theta)`. being in the same direction, add up [Fig.(b)]. Hence, the resultant intensity of electric field at the point P is
`E=E_(1) cos theta+E_(2) cos theta`
`=(1)/(4pi epsi_(0))(q)/((r^(2)+l^(2))) cos theta+(1)/(4pi epsi_(0))-(q)/((r^(2)+l^(2))) cos theta`
`=(1)/(4pi epsi_(0)) (q)/((r^(2)+l^(2))) 2 ocs theta`

Now, from figure (a),
`"cos" theta=(BO)/(BP)`
`=(l)/((r^(2)+l^(2))^(1//2))`
`:. E=(1)/(4pi epsi_(0))(q)/((r^(2)+l^(2)))(2l)/((r^(2)+l^(2))^(1//2))`
But 2ql=p (moment of electric dipole)
`:.E=(1)/(4pi epsi_(0))(p)/((r^(2)+l^(2))^(3//2))`
The direction of electric field E is .antiparallel to the dipole axis.
If r is very large compared to `2l (r gt gt 2l)`, then `l^(2)` may be neglected in comparison to `r^(2)`. Then, the electric field intensity at the point P due to the dipole is
`E=(1)/(4pi epsi_(0))(p)/(r^(3))` Newton//Coulomb
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