- #1
ks_wann
- 14
- 0
So, I've got a charge distribution given by:
\begin{equation}
\rho(r,\phi,z)=\frac{q}{2\pi R}\delta(r-R)\delta(z)\cos(2\phi)
\end{equation}
This, if I'm not mistaken, translates into a circular charge distribution located in the z-plane, a distance R from origo.
Thus
\begin{equation}
\rho(R,\phi,0)=\frac{q}{2\pi R}\cos(2\phi)
\end{equation}
I compute the total charge of the ring, which by my calculations is 0. Now I want to use
\begin{equation}
\mathbf{p}=\int\mathbf{r^{\prime}}\rho(\mathbf{r^{\prime}})dl^{\prime}
\end{equation}
to find the dipole moment, where
\begin{equation}
dl^{\prime}=Rd\phi.
\end{equation}
Inserting in the equation, I get:
\begin{equation}
\mathbf{p}=\int R\mathbf{\hat{r}}\frac{q}{2\pi R}\cos(2\phi)Rd\phi \\ = \frac{Rq}{2\pi}\mathbf{\hat{r}}\int_{0}^{2\pi} \cos(2\phi)d\phi = 0
\end{equation}
I'm not sure if the last steps makes sense..
I'd be grateful for any help.
\begin{equation}
\rho(r,\phi,z)=\frac{q}{2\pi R}\delta(r-R)\delta(z)\cos(2\phi)
\end{equation}
This, if I'm not mistaken, translates into a circular charge distribution located in the z-plane, a distance R from origo.
Thus
\begin{equation}
\rho(R,\phi,0)=\frac{q}{2\pi R}\cos(2\phi)
\end{equation}
I compute the total charge of the ring, which by my calculations is 0. Now I want to use
\begin{equation}
\mathbf{p}=\int\mathbf{r^{\prime}}\rho(\mathbf{r^{\prime}})dl^{\prime}
\end{equation}
to find the dipole moment, where
\begin{equation}
dl^{\prime}=Rd\phi.
\end{equation}
Inserting in the equation, I get:
\begin{equation}
\mathbf{p}=\int R\mathbf{\hat{r}}\frac{q}{2\pi R}\cos(2\phi)Rd\phi \\ = \frac{Rq}{2\pi}\mathbf{\hat{r}}\int_{0}^{2\pi} \cos(2\phi)d\phi = 0
\end{equation}
I'm not sure if the last steps makes sense..
I'd be grateful for any help.