Electric field inside a polarized sphere

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
4 replies · 5K views
Fabio010
Messages
84
Reaction score
0

Homework Statement



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

where k is a constant and [itex]\vec{r}[/itex] 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 -[itex]\frac{1}{3εo}[/itex]*[itex]\vec{P}[/itex]
??


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

In the exercise [itex]\vec{P}[/itex] is not uniform.
 
Physics news on Phys.org
Fabio010 said:
Can i consider always the electric field inside a sphere as -[itex]\frac{1}{3εo}[/itex]*[itex]\vec{P}[/itex]
??

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 [itex]\vec{E}[/itex](A) = [itex]\frac{∫-∇.\vec{P}dv}{εo}[/itex]
 
Last edited:
Fabio010 said:
So you are telling me that [itex]\vec{E}[/itex](A) = [itex]\frac{∫-∇.\vec{P}dv}{εo}[/itex]

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##.