Dipole Moment of a Ring

Consider a ring with radius $R$ and charge density $\lambda=\lambda_0\cos\phi$, where $\phi$ is the angular coordinate in the cylindrical coordinate. If I want to find the dipole moment of this charge distribution, then I put it into $$\vec{p}=\int{\mathrm d^3r~\rho(\vec{r})\vec{r}},$$ where I tried $$\rho(\vec{r})=\lambda_0 \cos\phi \times \delta(s-R) \delta(z)$$ and $$\vec{r}=s\hat{s}+z\hat{z}$$ So, the dipole then becomes: $$\vec{p}=\int_{-\infty}^{\infty}\int_{0}^{2\pi}\int_{0}^{\infty}[\lambda_0 \cos\phi \times \delta(s-R) \delta(z)](s\hat{s}+z\hat{z})s~\mathrm ds~\mathrm d\phi ~\mathrm dz$$ The delta function kills the $z$ component and leave the $s$ component, so: $$\vec{p}=\int_{-\infty}^{\infty}\int_{0}^{2\pi}\int_{0}^{\infty}\lambda_0 \cos\phi \times \delta(s-R) \delta(z)s^2~\mathrm ds~\mathrm d\phi ~\mathrm dz\hat{s}=\lambda_0R^2\int_{0}^{2\pi}\cos\phi~\mathrm d\phi~\hat{s}=\vec{0}$$ The answer is $$\vec{p}=\frac{1}{2}\lambda_0R^2\hat{x}$$ What's wrong with my solution? Actually this is problem 4.1 from Zangwill's Modern Electrodynamics. I read it's solution, but don't get why it works under Cartesian coordinates but not under cylindrical coordinate.

• I think the problem is your $\hat s$ is not constant in orientation as you go around the loop with $\phi$. – ZeroTheHero Feb 5 '17 at 21:31

The previous answer is just a repeat of whats in the solution manual and the p provided in this is incorrect. I am taking a class for grins and was assigned this problem and beat myself up trying to get the stated answer. I finally did it in both spherical and cylindrical coordinates and got the same answer in both. I emailed prof and he confirmed the answer in the book is incorrect. The actual answer is $\pi \lambda_0R^2\hat{x}$

note that the $\phi$ integration is $\int_0^{2\pi}cos^2 \phi = \pi$

the $\theta$ integral is $\int_0^{\pi} sin(\theta)\delta(\theta - \frac{\pi}{2})d\theta = sin(\frac{\pi}{2}) = 1$

the $r$ integral is $\int_0^{\infty} r^2\delta(r-R) = R^2$

so the answer is $\pi\lambda_0R^2\hat{x}$

For a linear charge density around a ring, $\lambda=\lambda(\phi)$. The volume charge density would be: $$\rho(r)=\lambda(\phi)\frac{\delta\left(\theta-\frac{\pi}{2}\right)}{\sin(\theta)}\frac{\delta(r-R)}{r}$$ Now the dipole moment would be: $$p=\int_{0}^{2\pi}~\mathrm d\phi\int_{0}^{\pi}~\sin(\theta)~\mathrm d\theta \int_{0}^{\infty}~\mathrm dr~r^{2}\lambda\frac{\delta\left(\theta-\frac{\pi}{2}\right)}{\sin(\theta)}\frac{\delta(r-R)}{r}=2\pi R\lambda=Q$$ Since $\hat{r}=\hat{z}\cos\theta +\hat{y}\sin\theta \sin\phi +\hat{x}\sin\theta \cos\phi$, the dipole moment would be: $$\int ~\mathrm d^{3}r~\rho(r) =\int_{0}^{2\pi}~\mathrm d\phi\int_{0}^{\pi}~\sin(\theta)~\mathrm d\theta \int_{0}^{\infty}~\mathrm dr~r^{2}~[\hat{z}\cos\theta +\hat{y}\sin\theta \sin\phi +\hat{x}\sin\theta \cos\phi]~\lambda_{0}\cos\phi\frac{\delta\left(\theta-\frac{\pi}{2}\right)}{\sin(\theta)}\frac{\delta(r-R)}{r}$$ The $\hat{x}$ integral is non zero and would result in $$\boxed{p=\frac{1}{2}\lambda_{0}R^{2}\hat{x}}$$

• Thanks for answering but this is the method used in Zangwill's solution manual. I want an explanation why my result is different from his. – LY3000 Nov 17 '16 at 11:53

protected by Qmechanic♦Feb 5 '17 at 19:52

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