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This is to be handled by the same trick. We have effectively a two dimensional situation. For a uniform cylinder full of charge with charge density $ρ_{0}$ (charge per unit volume), the electric field $E$ at an inside point is along the (cylindrical) radius vector $r$ and equal to,

$E=2ϵ_{0}1 ρr$

$(divE=rl ∂r∂ (rE_{r})=ϵ_{0}ρ ,hence,E_{r}=2ϵ_{0}ρ r)$

Therefore the polarized cylinder can be thought of as two equal and opposite charge distributions displaced with respect to each other

$E=2ϵ_{0}1 ρr−2ϵ_{0}1 ρ(r−δr)=2ϵ_{0}1 ρδr=−2ϵ_{0}P $

Since $P=−ρδr$. (direction of electric dipole moment vector being from the negative charge to positive charge.)

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