Physics

$p=q×2a$

where $p$ is the electric dipole moment pointing from the negative charge to the positive charge.

Electric field at A due to dipole, $E_{A}=r_{3}2KP $ where $K=4πϵ_{o}1 $

$E_{A}=(x+a)_{3}2KP $

Force at A due to dipole, $F_{A}=qE_{A}=(λ$ $dx)$ $×(x+a)_{3}2KP $

$F_{A}=(x+a)_{3}2λKP dx$

Total force exerted due to dipole on the rod, $F_{D}=∫_{0}(x+a)_{3}2λKP dx$

$F_{D}=2λKP×−2(x+a)_{2}1 ∣∣∣∣∣ _{0}$

$⟹F_{D}=a_{2}λKP $

Now force exerted by rod on dipole is equal and opposite to $F_{D}$ (Newton's law)

$F_{R} =−F_{D}=4πϵ_{o}a_{2}−λP $

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