Electric field inside a polarized sphere

In summary, the conversation revolves around finding the field inside and outside a sphere with a polarization of \vec{P}= k\vec{r}, where \vec{r} is the vector from the center and k is a constant. The field outside the sphere is determined to be zero due to the sum of the polarization charges being zero. The main question is whether the electric field inside the sphere can always be considered as -\frac{1}{3εo}*\vec{P} and the conversation concludes that it is not always the case, as the polarization charges inside the sphere create a bound charge density that can be used to calculate the field using Gauss' law.
  • #1
Fabio010
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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.
 
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  • #2
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.
 
  • #3
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.
 
  • #4
So you are telling me that [itex]\vec{E}[/itex](A) = [itex]\frac{∫-∇.\vec{P}dv}{εo}[/itex]
 
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  • #5
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##.
 

FAQ: Electric field inside a polarized sphere

1. What is an electric field inside a polarized sphere?

The electric field inside a polarized sphere is the force exerted on a charged particle inside the sphere due to the presence of the polarization of the material that makes up the sphere.

2. How is the electric field inside a polarized sphere calculated?

The electric field inside a polarized sphere can be calculated using the formula E = kP/r^3, where E is the electric field, k is the Coulomb constant, P is the polarization of the material, and r is the distance from the center of the sphere.

3. Does the electric field inside a polarized sphere vary with distance?

Yes, the electric field inside a polarized sphere does vary with distance. As the distance from the center of the sphere increases, the electric field decreases due to the inverse square law.

4. How does the polarization of the sphere affect the electric field inside?

The polarization of the sphere directly affects the magnitude and direction of the electric field inside. A higher polarization leads to a stronger electric field, while a lower polarization leads to a weaker electric field.

5. Can the electric field inside a polarized sphere be negative?

Yes, the electric field inside a polarized sphere can be negative. This occurs when the polarization of the material is directed in the opposite direction of the electric field, resulting in a cancellation of the field.

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