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

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1. May 10, 2015

### Aristotle

1. The problem statement, all variables and given/known data

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

3. 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!

2. May 10, 2015

### Aristotle

Bump

3. May 11, 2015

### haruspex

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. May 11, 2015

### Aristotle

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. May 11, 2015

### haruspex

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. May 11, 2015

### nrqed

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.