How do I go from an Electric Field graph to Charge Density?

Click For Summary
SUMMARY

This discussion focuses on the transition from an Electric Field graph to Charge Density using fundamental equations such as E = kq/(r^2) and E*dA = Q/ε₀. Participants analyze the behavior of electric fields in different materials, identifying regions as insulators, conductors, or space based on the graph's characteristics. Key insights include the application of Gauss' Law to relate electric field behavior to charge density, emphasizing that the electric field in a conductor is zero at equilibrium and how to interpret the electric field's dependence on radius.

PREREQUISITES
  • Understanding of Electric Field concepts and equations
  • Familiarity with Gauss' Law and its applications
  • Knowledge of charge distributions in conductors and insulators
  • Ability to interpret graphical representations of electric fields
NEXT STEPS
  • Study Gauss' Law in detail, focusing on its application to spherical charge distributions
  • Learn about electric field behavior in different materials, specifically insulators and conductors
  • Explore the mathematical relationship between electric fields and charge density
  • Investigate graphical analysis techniques for interpreting electric field and potential graphs
USEFUL FOR

Students and educators in physics, particularly those studying electromagnetism, as well as anyone involved in electrical engineering or related fields seeking to deepen their understanding of electric fields and charge distributions.

Aristotle
Messages
169
Reaction score
1

Homework Statement


Screen shot 2015-05-10 at 6.51.54 AM.png


Homework Equations


E= kq/(r^2), E*dA = Q/e0

The Attempt at a Solution


Typically I understand how to interpret basic graphs such as going for V (potential) vs x graph to Electric field vs x graph by finding the slope of V since E= -grad V...and from their it's basic algebra.
As for going from E vs r to charge density, I'm a bit lost on how to do so for #2.

Is there a formula to follow such as what I described above in interpreting these graphs and possibly something to help me see the big picture?
My attempt at this problem is...and correct me if I am wrong...
From 0-a it seems that the electric field of the spherical object is increasing, so I'd say it's an insulator. From a-b there is a drop of 1/r graph...so imagining that we are far away the e-field would approach zero, so I'd say it would be "space"---that is what they meant by space right?--(nothing present/no material)

Thank you for your help!
 
Physics news on Phys.org
Bump
 
I think you are on the right lines, though I'm not sure whether you are supposed to provide all feasibly correct answers. Is there any curve that could not be the result of a charge distribution on an insulator?
The section from a to b doesn't look all that much like 1/r to me, but it might not be well drawn.
What do you propose beyond b?
 
haruspex said:
I think you are on the right lines, though I'm not sure whether you are supposed to provide all feasibly correct answers. Is there any curve that could not be the result of a charge distribution on an insulator?
The section from a to b doesn't look all that much like 1/r to me, but it might not be well drawn.
What do you propose beyond b?
I see that at b-c, it is at zero so then it must be a conductor since e field is 0 in conductors for stationary charges. And from c-d i assume it to be space? Since the problem states that no two connecting region are made from same material
 
Aristotle said:
I see that at b-c, it is at zero so then it must be a conductor since e field is 0 in conductors for stationary charges. And from c-d i assume it to be space? Since the problem states that no two connecting region are made from same material
As I said, I don't see why you can't replace any layer with a suitably charged (or uncharged) insulator, provided no adjacent layer is an insulator.
0 to a must be an insulator, so a to b must be space. But b to c could be either a conductor, or an insulator with exactly the same charge distribution as the conductor would have. c to d could be anything other than whatever you choose for b to c.
 
Aristotle said:

Homework Statement


View attachment 83286

Homework Equations


E= kq/(r^2), E*dA = Q/e0

The Attempt at a Solution


Typically I understand how to interpret basic graphs such as going for V (potential) vs x graph to Electric field vs x graph by finding the slope of V since E= -grad V...and from their it's basic algebra.
As for going from E vs r to charge density, I'm a bit lost on how to do so for #2.

Is there a formula to follow such as what I described above in interpreting these graphs and possibly something to help me see the big picture?
My attempt at this problem is...and correct me if I am wrong...
From 0-a it seems that the electric field of the spherical object is increasing, so I'd say it's an insulator. From a-b there is a drop of 1/r graph...so imagining that we are far away the e-field would approach zero, so I'd say it would be "space"---that is what they meant by space right?--(nothing present/no material)

Thank you for your help!
Hint: You know the E field in a conductor is zero (at equilibrium). In an insulator with a spherically symmetric charge distribution, how does the E field behave as a function of radius? And in space around a spherical charge distribution, how does the E field behave?
As for the charge density, recall Gauss' law: the net electric flux through a gaussian surface is equal to the net charge enclosed over epsilon_0. And the net charge enclosed is the integral of the charge density. Consider then an electric field whose magnitude depends on r only, E= E(r). What is the flux through a spherical gaussian surface? Now, write this as \int 4 \pi \rho(r) r^2 dr / \epsilon_o. Then, if you know how E depends on r you can figure out how \rho depends on r.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
960
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
11
Views
1K
  • · Replies 26 ·
Replies
26
Views
3K
Replies
10
Views
2K
Replies
1
Views
2K
Replies
4
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K