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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Related: https://physics.stackexchange.com/questions/8221/why-isnt-there-a-centre-of-charge be links therein – Emilio Pisanty Apr 17 '21 at 23:30
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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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@Elis Lovelace In typical usage, "Dipole moment" would exclude the DC or monopole component and would be independent of the total charge. i.e. the computation would be taken around the "center of charge" at h=0. – Roger Wood Apr 18 '21 at 03:56