If we consider a cylinder along the $z$-axis, with charge density $\rho(z)$ [varying linearly according to$\rho = az + b$, from the positive value $\rho_1=-ha+b$] at $-h$ to the also positive $\rho_2=ah+b$ at $h$), the total charge is $Q\neq0$. In that case, how can we find direction of the dipole moment? Because of rotation symmetry, $\vec{p}$ should be oriented along $z$-axis $\vec{p} = (0,0,p_z)$. Can we say, that the vector points in the direction of a place with higher charge density? I think that if total charge $Q\neq0$, the result depends on the coordinate origin.
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You are right, since the total charge is not zero, the dipole moment depends on the point of reference:
The dipole moment with respect to the point at $\vec{x}_0$ is
$$\vec{p}_{x_0}=\int d^3\vec{x}\,\rho(\vec{x}+\vec{x}_0)\,\vec{x}\,\,\,.\tag{1}$$
Performing a change of variable $\vec{x}+\vec{x}_0=\vec{y}$ in the integral,
$$\vec{p}_{x_0}=\int d^3\vec{y}\,\rho(\vec{y})\,(\vec{y}-\vec{x}_0)=\int d^3\vec{y}\,\rho(\vec{y})\,\vec{y}-\vec{x}_0\int d^3\vec{y}\,\rho(\vec{y})=\vec{p}_{origin}-Q\vec{x}_0,$$
where $Q$ is the total charge.
AFG
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