It is very important to understand the mathematical beauty of the terms occurring in the Multipole expansion. Since higher order terms vanish faster than the monopole, dipole terms, they are important only when the need of the accuracy is high.
As a consequence, in most of the common problems, dipole terms plays the most significant role after the monopole term .
The dipole term in the expansion is ,
$$ V(\vec{r})_{dipole} = \frac{1}{4\pi\epsilon_0} \frac{1}{r^2} \int_{V'} r'cos\theta' \rho(\vec{r'}) \,dV' ...\ldots eq.(1)$$
To make the integrand a vector quantity, we know that
$\hat{r} \cdot \vec{r'} = r' cos\theta' $
and so the dipole term becomes, $$ V(\vec{r})_{dipole} = \frac{\hat{r}}{4\pi\epsilon_0 r^2}\cdot \int_{V'} \vec{r'} \rho(\vec{r'}) \,dV'$$ Terms they depend only on r' is separated and called as dipole moment, $$ \vec{p} = \int_{V'} \vec{r'} \rho(\vec{r'}) \,dV' ...\ldots eq.(2)$$
and the dipole potential becomes, $$ V_{dipole}(\vec{r}) = \frac {\vec{p}\cdot \hat{r}}{4\pi\epsilon_0 r^2} ...\ldots eq.(3)$$ From eq.(2) it is known that the dipole moment defined only in terms of r' and it implies that, 'dipole moment of a volume charge depends only on it distribution'.
For point charges, $$ \vec{p} = \sum_{i=1}^N q_i \vec{r'_i}$$ and for a physical dipole it becomes, $ \vec{p} = q(\vec{r'_+} - \vec{r'_-}) = q\vec{d}$ where d is the vector from -q to +q.
But we should remember that dipole moment doesn't mean there should be only two charges. Our definition is general for any charge distribution. It so happens the physical dipole has similar kind of representation. It is always possible to ask for the dipole moment of any number of charges e.g. three charges in a triangle.
Some properties of the dipole moment are, Change in coordinate system usually changes the dipole moment except when the total charge is zero. And if we place a physical dipole in a uniform Electric field E, it will experiences a torque and if it is non-uniform it will experience an additional force other than the torque given by, $ \vec{F} = (\vec{p}\cdot\nabla)\vec{E} $
Each problem will give more insight.
Similar kind of multipole expansion for a vector potential reveals that the dipole term for a vector potential as,
$$ \vec{A_{dipole}(\vec{r})} = \frac{\mu_0}{4\pi} \frac {\vec{m}\times\hat{r}}{r^2} $$ where $\vec{m}$ is the magnetic dipole moment $$ \vec{m} = I \int \,d{\vec{a}} = I\vec{a} $$
'a' is the area enclosed by the loop and 'I' is the current. Magnetic dipole moment is always independent of the coordinate system since they don't play any role.
As we did in Electric dipole, certain properties for a magnetic dipole are obtained and are,
In a uniform field, the net force on any loop is zero. In a non-uniform magnetic field, an infinitesimal loop of dipole moment 'm' will experience a force, $$ \vec{F} = \nabla (\vec{m}\cdot\vec{B})$$ Thus, the concept of polarization explains the newer ideas namely, bound charges and bound currents.
As a consequence, in most of the common problems, dipole terms plays the most significant role after the monopole term .
The dipole term in the expansion is ,
$$ V(\vec{r})_{dipole} = \frac{1}{4\pi\epsilon_0} \frac{1}{r^2} \int_{V'} r'cos\theta' \rho(\vec{r'}) \,dV' ...\ldots eq.(1)$$
To make the integrand a vector quantity, we know that
$\hat{r} \cdot \vec{r'} = r' cos\theta' $
and so the dipole term becomes, $$ V(\vec{r})_{dipole} = \frac{\hat{r}}{4\pi\epsilon_0 r^2}\cdot \int_{V'} \vec{r'} \rho(\vec{r'}) \,dV'$$ Terms they depend only on r' is separated and called as dipole moment, $$ \vec{p} = \int_{V'} \vec{r'} \rho(\vec{r'}) \,dV' ...\ldots eq.(2)$$
and the dipole potential becomes, $$ V_{dipole}(\vec{r}) = \frac {\vec{p}\cdot \hat{r}}{4\pi\epsilon_0 r^2} ...\ldots eq.(3)$$ From eq.(2) it is known that the dipole moment defined only in terms of r' and it implies that, 'dipole moment of a volume charge depends only on it distribution'.
For point charges, $$ \vec{p} = \sum_{i=1}^N q_i \vec{r'_i}$$ and for a physical dipole it becomes, $ \vec{p} = q(\vec{r'_+} - \vec{r'_-}) = q\vec{d}$ where d is the vector from -q to +q.
But we should remember that dipole moment doesn't mean there should be only two charges. Our definition is general for any charge distribution. It so happens the physical dipole has similar kind of representation. It is always possible to ask for the dipole moment of any number of charges e.g. three charges in a triangle.
Some properties of the dipole moment are, Change in coordinate system usually changes the dipole moment except when the total charge is zero. And if we place a physical dipole in a uniform Electric field E, it will experiences a torque and if it is non-uniform it will experience an additional force other than the torque given by, $ \vec{F} = (\vec{p}\cdot\nabla)\vec{E} $
Each problem will give more insight.
Similar kind of multipole expansion for a vector potential reveals that the dipole term for a vector potential as,
$$ \vec{A_{dipole}(\vec{r})} = \frac{\mu_0}{4\pi} \frac {\vec{m}\times\hat{r}}{r^2} $$ where $\vec{m}$ is the magnetic dipole moment $$ \vec{m} = I \int \,d{\vec{a}} = I\vec{a} $$
'a' is the area enclosed by the loop and 'I' is the current. Magnetic dipole moment is always independent of the coordinate system since they don't play any role.
As we did in Electric dipole, certain properties for a magnetic dipole are obtained and are,
In a uniform field, the net force on any loop is zero. In a non-uniform magnetic field, an infinitesimal loop of dipole moment 'm' will experience a force, $$ \vec{F} = \nabla (\vec{m}\cdot\vec{B})$$ Thus, the concept of polarization explains the newer ideas namely, bound charges and bound currents.
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