Find ELectric field in these regions of a spherical shell

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SUMMARY

The discussion centers on calculating the electric field (E) in three regions of a thick spherical shell with a charge density of k/r², specifically for the regions r PREREQUISITES

  • Understanding of Gauss's Law (E dot da = Q/ε)
  • Familiarity with electric fields in spherical charge distributions
  • Knowledge of integration techniques in physics
  • Concept of charge density and its implications
NEXT STEPS
  • Study the implications of Gauss's Law in electrostatics
  • Learn about electric fields produced by different charge distributions
  • Explore the concept of charge density in various geometries
  • Review integration techniques specific to physics problems
USEFUL FOR

Students and educators in physics, particularly those focusing on electromagnetism, as well as anyone seeking to understand electric fields in spherical charge distributions.

grandpa2390
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Homework Statement


a thick spherical shell carries charge density k/r^2 a<r<b
find E in the three regions r<a a<r<b b<r

Homework Equations


E dot da = Q/ε

The Attempt at a Solution


I can't understand why, when integrating, they choose for
ii to integrate between a and r,
iii and the between a and b for iii
 
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grandpa2390 said:

Homework Statement


a thick spherical shell carries charge density k/r^2 a<r<b
find E in the three regions r<a a<r<b b<r

Homework Equations


E dot da = Q/ε

The Attempt at a Solution


I can't understand why, when integrating, they choose for
ii to integrate between a and r,
iii and the between a and b for iii
What do you know about the field inside a uniformly charged spherical shell?
 
haruspex said:
What do you know about the field inside a uniformly charged spherical shell?
that it's uniform at the surface?
I don't know what you are asking.

what I have been able to gather is that they are integrating the volume of the shell. integral of 4 pi r^2 dr
for a<r<b the volume is from a - r
for b<r the volume is from a-b

I don't know why.
 
grandpa2390 said:
that it's uniform at the surface?
No, that there is no field produced inside a uniformly charged spherical shell. This is a fundamental result of enormous importance in these problems. The equally important result for outside the shell is that the field there is the same as if all of the charge were concentrated at the sphere's centre.
The same pair of results applies (of course) to gravitational fields from uniform spherical mass distributions.

Can you see how this explains the integration range?
 
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haruspex said:
No, that there is no field produced inside a uniformly charged spherical shell. This is a fundamental result of enormous importance in these problems. The equally important result for outside the shell is that the field there is the same as if all of the charge were concentrated at the sphere's centre.
The same pair of results applies (of course) to gravitational fields from uniform spherical mass distributions.

Can you see how this explains the integration range?

We are trying to capture all of the "mass" below our boundary.

Between A and B we want to capture all the mass from a to wherever r is.
If R is greater then B then we want all of the "mass" less then r which is from a to b

?
 
grandpa2390 said:
Between A and B we want to capture all the mass from a to wherever r is.
Hence the integration range from a to r.
grandpa2390 said:
If R is greater then B then we want all of the "mass" less then r which is from a to b
Hence the integration range from a to b.
You seem to have answered your own questions.
Please try to explain more clearly what it is that you do not understand.
 
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haruspex said:
Hence the integration range from a to r.

Hence the integration range from a to b.
You seem to have answered your own questions.
Please try to explain more clearly what it is that you do not understand.

No you answered it. Or at the very least, you slapped some sense into my brain, pointed... pushed my brain into the right direction.
I don't know. When you compared it to mass, it just made sense to me suddenly. I don't know. I was thinking it should have been integrated between the boundaries stated. integrated from a to b, and then from b to infinity. I didn't get it until your last reply :)

Then I just restated what I got from you in my own words for verification to make sure whether I had it : )

Thanks for your help!
 
grandpa2390 said:
restated what I got from you in my own words for verification
Verified.
 
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