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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/(4 pi epsilon_0)) p/((r^2 + l^2)^(3//2))`
Where the terms have their usual meaning.
Show that intensity of electric field E at a point in broadside on position is given by:
E = `(1/(4 pi epsilon_0)) p/((r^2 + l^2)^(3//2))`,
where the terms have their usual meaning.
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उत्तर
Consider a dipole of length 2l and moment `vec p`.
From the figure, resultant electric field intensity at C:
`vec E = vec E_A + vec E_B`
`|vec E_A| = 1/(4 pi epsilon_0) q/(r^2 + l^2)`
`|vec E_B| = 1/(4 pi epsilon_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/(4 pi epsilon_0) * q/(r^2 + l^2) cos theta + 1/(4 pi epsilon_0) * q/(r^2 + l^2) cos theta`
= `2 * 1/(4 pi epsilon_0) * q/(r^2 + l^2) cos theta`
= `2 xx 1/(4 pi epsilon_0) * q/(r^2 + l^2) xx l/(r^2 + l^2)^(1//2)`
= `1/(4 pi epsilon_0) (q(2 l))/((r^2 + l^2)^(3//2))`
E = `1/(4 pi epsilon_0) p/((r^2 + l^2)^(3//2))` ...[as p = q(2l)]
The above expression gives the magnitude of the field. The direction of the electric field E at C is opposite to the direction of the dipole moment `vec p`.
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