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

In summary, the conversation discusses interpreting graphs and determining the behavior of electric fields and charge density in different materials and scenarios. It mentions using formulas and laws such as Gauss' law and the equation for electric field, as well as considering the symmetry of a charge distribution. The main topics covered include identifying insulators and conductors, understanding the behavior of electric fields in space and around a spherical charge distribution, and using Gauss' law to determine charge density.
  • #1
Aristotle
169
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!
 
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  • #2
Bump
 
  • #3
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?
 
  • #4
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
 
  • #5
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.
 
  • #6
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 [itex] \int 4 \pi \rho(r) r^2 dr / \epsilon_o [/itex]. Then, if you know how E depends on r you can figure out how [itex] \rho [/itex] depends on r.
 

1. How do I determine the charge density from an electric field graph?

The charge density can be determined by calculating the slope of the electric field graph at any point. The slope represents the electric field strength at that point, which is directly proportional to the charge density.

2. What does the shape of an electric field graph tell us about the charge density?

The shape of an electric field graph can tell us the distribution of charge density. A steeper slope indicates a higher charge density, while a flatter slope indicates a lower charge density. A flat line indicates a uniform charge density, while a curved line suggests a non-uniform distribution.

3. Can the electric field graph accurately determine the charge density of a system?

Yes, the electric field graph can accurately determine the charge density of a system. However, it is important to note that the electric field graph only provides a qualitative representation of the charge density and is not a precise measurement.

4. How does the distance between charges affect the electric field graph and charge density?

The distance between charges directly affects the electric field graph and the charge density. As the distance between charges increases, the electric field strength decreases, resulting in a flatter slope on the graph and a lower charge density. As the distance between charges decreases, the electric field strength increases, resulting in a steeper slope on the graph and a higher charge density.

5. Can the electric field graph be used to calculate the total charge of a system?

No, the electric field graph cannot be used to directly calculate the total charge of a system. It only provides information about the distribution and strength of the electric field, which is related to the charge density. To calculate the total charge, other methods such as Gauss's law or Coulomb's law must be used.

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