How Do Charge Distributions Affect a Spherical Dielectric Shell?

Click For Summary
SUMMARY

The discussion focuses on the effects of charge distributions within a spherical dielectric shell, specifically when a charge Q is placed at the center. The solutions for the polarization P in the region a < r < b and the volume charge density ρ_pv are established, with ρ_pv equating to zero due to the divergence of P. The surface charge densities ρ_ps at the boundaries r = a and r = b are derived as -Q/(4*pi*a^2) * (ε_r-1)/ε_r and -Q/(4*pi*b^2) * (ε_r-1)/ε_r, respectively. The negative signs in these expressions are attributed to the orientation of the normal vectors at the surfaces.

PREREQUISITES
  • Understanding of dielectric materials and their properties
  • Familiarity with Maxwell's equations, particularly Gauss's law
  • Knowledge of vector calculus, specifically divergence and surface integrals
  • Basic concepts of electrostatics, including charge distributions and polarization
NEXT STEPS
  • Study the derivation of Gauss's law for different geometries
  • Learn about the behavior of electric fields in dielectric materials
  • Explore the mathematical techniques for calculating divergence in vector fields
  • Investigate the implications of surface charge densities in electrostatic problems
USEFUL FOR

Students and professionals in physics and electrical engineering, particularly those focusing on electrostatics and dielectric materials, will benefit from this discussion.

fkf
Messages
9
Reaction score
0

Homework Statement


Consider a spherical dielectric shell so that ε = ε_0ε_r for a < r < b and ε = ε_0 for 0 < r < a. If a charge Q is placed at the center of the shell, find

a) P for a < r < b
b) ρ_pv for a < r b
c) ρ_ps at r = a and r = b

Homework Equations


ρ_pv = -div(P)
ρ_ps = P \cdota_n


The Attempt at a Solution


I've managed to solve a) with the answer
Q/(4*pi*r^2) * (ε_r-1)/ε_r​
which is correct (checked the answers).

I've also solved b) which is ρ_pv = 0 since ρ_pv = -div(P).

I have trouble to find ρ_ps at r = a and r = b. The answer states that
-Q/(4*pi*a^2) * (ε_r-1)/ε_r and -Q/(4*pi*b^2) * (ε_r-1)/ε_r​
respectively. Where is that negative sign coming from?
 
Physics news on Phys.org
On the r = a I would guess that we get a negative sign since we're having a unit vector a_r which goes outward from the origin. And since the normal to the surface when r = a is the -a_r I would guess that the negative sign when r = a is correct. But when r = b, the unit vector from P and a_r are booth positive, so why negative there?
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
5K
  • · Replies 1 ·
Replies
1
Views
903
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 29 ·
Replies
29
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
2
Views
2K
  • · Replies 13 ·
Replies
13
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
4
Views
5K