Electric field inside a polarized sphere

Fabio010
Messages
84
Reaction score
0

Homework Statement



A sphere of radius R carries a polarization
\vec{P}= k\vec{r},

where k is a constant and \vec{r} is the vector from the center.


Find the field inside and outside the sphere.


In solution, the field outside sphere is 0.

I interpreted that as the field produced by the polarization charges.

Their sum is 0. So, the outside field is zero.

My problem is inside sphere.

Can i consider always the electric field inside a sphere as -\frac{1}{3εo}*\vec{P}
??


Because the formula -\frac{1}{3εo}*\vec{P} is obtained by a uniformly polarized sphere.

In the exercise \vec{P} is not uniform.
 
Physics news on Phys.org
Since there is no free charge anywhere inside the sphere, I would think the E field, by Gauss, would be zero everywhere inside the sphere.
 
Fabio010 said:
Can i consider always the electric field inside a sphere as -\frac{1}{3εo}*\vec{P}
??

Not sure where you're getting that expression. The polarization produces a bound charge density inside the sphere: ##\rho_b = -\vec{\nabla}\cdot \vec{P}##

From the bound charge you can get the field using Gauss' law.
 
So you are telling me that \vec{E}(A) = \frac{∫-∇.\vec{P}dv}{εo}
 
Last edited:
Fabio010 said:
So you are telling me that \vec{E}(A) = \frac{∫-∇.\vec{P}dv}{εo}

Yes (although I'm not real clear on your notation in this equation). See what you get for the bound charge density as a function of ##r## and then use it in Gauss' law to find the magnitude of the field inside, ##E(r)##, as a function of ##r##.
 
##|\Psi|^2=\frac{1}{\sqrt{\pi b^2}}\exp(\frac{-(x-x_0)^2}{b^2}).## ##\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).## ##y=x-x_0 \quad x=y+x_0 \quad dy=dx.## The boundaries remain infinite, I believe. ##\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dy(y+x_0)\exp(\frac{-y^2}{b^2}).## ##\frac{2}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,y\exp(\frac{-y^2}{b^2})+\frac{2x_0}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,\exp(-\frac{y^2}{b^2}).## I then resolved the two...
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