Electric field inside a uniformly charged insulator

Click For Summary
SUMMARY

The discussion focuses on calculating the electric field inside a uniformly charged insulating sphere using Gauss's Law. The key equation for this scenario is φ = q_encl/ε₀, where ε₀ is the permittivity of free space. It is established that any point inside a spherical shell experiences a net electric field of 0 N/C, and only the charge within the radial distance contributes to the electric field. The confusion arises regarding whether to use ε₀ or the material's permittivity (κ) when the insulating sphere is made of a material other than air.

PREREQUISITES
  • Understanding of Gauss's Law in electrostatics
  • Knowledge of electric field concepts and vector analysis
  • Familiarity with permittivity constants (ε₀ and κ)
  • Basic principles of charge distribution in spherical geometries
NEXT STEPS
  • Study the application of Gauss's Law in different geometries
  • Learn about electric field calculations in non-uniform charge distributions
  • Explore the differences between electric field strength (E) and electric flux density (D)
  • Investigate the implications of material properties on electric fields
USEFUL FOR

Students studying electrostatics, physics educators, and anyone interested in understanding electric fields in charged insulating materials.

0blivi0n
Messages
4
Reaction score
0

Homework Statement


ok here's the problem: find the magnitude of the electric field inside uniformly charged insulating sphere of raduis R.


Homework Equations


application of gauss's law..but...


The Attempt at a Solution


should i use
\phi = q_{encl}\epsilon_0
or
\phi = q_{encl}/\kappa\epsilon
?
 
Physics news on Phys.org
It becomes slightly different inside a sphere. Electric fields are vectors, and a point inside a sphere is being affected by an electric field in every direction. Heres what you need to know: Any point inside a spherical shell of any thickness receives a net electric field of 0N/C. Same concept applies to electric force and gravity.
Here is a site to help you out on this:
http://hyperphysics.phy-astr.gsu.edu/Hbase/electric/elesph.html
Using that, if a point is a radial distance a from the center of the sphere (while inside the sphere), the net electric field at a point inside a sphere would be due to only the the charge within the radial distance (the charges outside the radial distance contributes to the 0N/C).
 
Last edited:
So, are you in space or a material? Would you want to use the electric constant for free space or not?
 
Mindscrape said:
So, are you in space or a material? Would you want to use the electric constant for free space or not?
that is what I'm confuswed about...if the insulating sphere is made of a material other than air, which one should I use on the righthand side of the flux equation? \epsilon_{0} or \epsilon. that is should I take the permittivity constant (\kappa) into account or not?
 
Since you are solving for the electric field strength E, you would just use \epsilon_{o}. If you were asked for the electric flux density D, you would need to be concerned with the electric permittivity of the material. (The hint is that you aren't given a value for \kappa in the problem...)
 
0blivi0n said:

The Attempt at a Solution


should i use
\phi = q_{encl}\epsilon_0
or
\phi = q_{encl}/\kappa\epsilon
?

Out of curiosity, what volume are you using for the enclosed charge?
 
a sphere
 
thanks dynamicsolo. I'm starting to see how things are now
 
0blivi0n said:
a sphere

thats a shape, not a volume
 
  • #10
If a charge were distributed uniformly on the surface of the balloon(insulator). A point particle with charge q inside is greatest when it is anywhere inside the sphere because the force is zero?

or when it is near the inside surface of the balloon?
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
968
Replies
4
Views
5K
Replies
23
Views
4K
  • · Replies 26 ·
Replies
26
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 6 ·
Replies
6
Views
4K
Replies
9
Views
829
  • · Replies 3 ·
Replies
3
Views
2K