- #1
LeoJakob
- 21
- 1
- Homework Statement
- Consider the non-uniformly charged hollow sphere with radius ##R## and charge density##
\rho(\vec{r})=\rho_{0} \cos \theta \delta(r-R) .##
Calculate the dipole moment ##\vec{p}##
- Relevant Equations
- ##
\vec{p}=\int \vec{r} \rho(\vec{r})\mathrm{d}^{3} r .
##
is there an easier way to calculate the dipole moment? I described ## \vec r## in spherical coordinates. I thought at first that due to the symmetry I can assume that dipole-moment only points in the ##z##-direction, but the charge distribution is inhomogeneous, so I made the following calculation:
My calculation results in $$\vec{p}=\int \vec{r} \rho(\vec{r}) \mathrm{d}^{3} r=\int \limits_{0}^{2 \pi} \int \limits_{0}^{\pi} \int \limits_{0}^{\infty} R\left(\sin \theta \cos \phi \vec{e}_{x}+\sin \theta \sin \phi \vec{e}_{y}+\cos \theta \vec{e}_{z}\right)r^{2} \cdot \delta(r-R) \rho_{0} \sin \theta \cos \theta d r d \theta d \phi =\frac{4}{3} \rho_{0} R^{3} \overrightarrow{e_{z}}$$
Which is the correct result but the calculation of the integrals took quite some time. In the end I realized that the dipole moment,indeed, only has a ##z##-component, could I have recognized this earlier and thus simplified my calculation? I'm unsure because the charge density is not homogenous.
My calculation results in $$\vec{p}=\int \vec{r} \rho(\vec{r}) \mathrm{d}^{3} r=\int \limits_{0}^{2 \pi} \int \limits_{0}^{\pi} \int \limits_{0}^{\infty} R\left(\sin \theta \cos \phi \vec{e}_{x}+\sin \theta \sin \phi \vec{e}_{y}+\cos \theta \vec{e}_{z}\right)r^{2} \cdot \delta(r-R) \rho_{0} \sin \theta \cos \theta d r d \theta d \phi =\frac{4}{3} \rho_{0} R^{3} \overrightarrow{e_{z}}$$
Which is the correct result but the calculation of the integrals took quite some time. In the end I realized that the dipole moment,indeed, only has a ##z##-component, could I have recognized this earlier and thus simplified my calculation? I'm unsure because the charge density is not homogenous.