Dipole moment of non-conducting spherical shell

jackxxny
Messages
39
Reaction score
0

Homework Statement


I'm trying to find the dipole moment of

The surface charge distribution is:

<br /> \sigma = \sigma_{0} sin 2 \theta <br />

Homework Equations


The Attempt at a Solution



<br /> p_z=\int_{0}^{\pi}{\sigma* z* dA } = \int_{0}^{\pi}{(\sigma_0 \sin{2 \theta})*(acos{\theta})*(2\pi a^2 \sin{\theta}d\theta)}<br />

Doing so I obtain

<br /> ( \sigma_{0} a^3 \pi^2)/(2)<br /> <br />

Can that be the answer?
 
Physics news on Phys.org
You should also explicitly show (or give a good argument for why) p_x=p_y=0 and then write your final answer in vector form (since dipole moment is a vector!) as \textbf{p}=\frac{\sigma_{0} a^3 \pi^2}{2}\mathbf{\hat{z}}, but other than that it looks good to me!:approve:
 
Hello everyone, I’m considering a point charge q that oscillates harmonically about the origin along the z-axis, e.g. $$z_{q}(t)= A\sin(wt)$$ In a strongly simplified / quasi-instantaneous approximation I ignore retardation and take the electric field at the position ##r=(x,y,z)## simply to be the “Coulomb field at the charge’s instantaneous position”: $$E(r,t)=\frac{q}{4\pi\varepsilon_{0}}\frac{r-r_{q}(t)}{||r-r_{q}(t)||^{3}}$$ with $$r_{q}(t)=(0,0,z_{q}(t))$$ (I’m aware this isn’t...
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
Back
Top