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Charge density in a resistor issue

  1. Sep 9, 2011 #1
    1. The problem statement, all variables and given/known data

    given simple circuit with V the voltage connected to a resistor
    with area A, length d and electrical conductance: b*e^-2x/d
    after my computations the resistance

    R=d*(e^2-1)/(2bA)


    2. Relevant equations

    what is the electrical field E in the center of the resistor?

    3. The attempt at a solution

    i think it must be 0 because i can't think of
    why it will be charged
     
  2. jcsd
  3. Sep 9, 2011 #2
    Hey,
    Since you didn't provide any formulae, as to how you'd reached your result, let me ask you this: are you familiar with the relation: [tex] \sigma E = j [/tex]?(Local variation of Ohm's law) Where sigma is the conductance, and j is the current density.
    Since the voltage is constant, E must be uniform, simple integration should yield what you're looking for...
    I hope that helps somewhat,
    Daniel
    P.S
    Clearly, by following the steps above, you can find that the density won't be zero, per se
    The "continuity equation", however, if you recall, is still satisfied(naturally), and generates that equilibrium we all seek...
     
    Last edited: Sep 9, 2011
  4. Sep 9, 2011 #3
    yeah..oops, this is the second question and I managed that one
    I know the formula :

    [tex] \sigma E = j [/tex](Local variation of Ohm's law)

    the 3rd question was the charge density
    in the center of the resistor

    that i couldn't figure out..
     
  5. Sep 9, 2011 #4
    Have you tried Maxwell's equation->[tex] \overrightarrow{\nabla} \cdot \overrightarrow{E} = \frac{\rho}{\epsilon_0} [/tex]
    Does that get you any closer?
    Daniel
     
  6. Sep 9, 2011 #5
    becasue E is E(x), can i say that

    [tex] \overrightarrow{\nabla} \cdot \overrightarrow{E} = \frac{\rho}{\epsilon_0}[/tex]
    is
    [tex] \frac{dE(x)}{dx} = \frac{\rho(x)}{\epsilon_0} [/tex]

    ?
    then problem is solved
     
  7. Sep 9, 2011 #6
    Absolutely! But of course, this could only work by stating that the area(A) is also only dependant on x... so long as that's maintained, you're prefectly correct.
    Did you end-up reaching the desired answer?
    I hope was I able to provide some aid,
    Daniel
     
  8. Sep 10, 2011 #7
    it doesn't say what is the correct answer but it satisfy me.

    so any resistor is also charged (when connected to voltage)? sound weird to me
    because out of symmetry why it would be charged positively
    and not the opposite?
     
  9. Sep 10, 2011 #8
    Since the current travels via a relative value known as the "drift velocity", the charges that encounter the most resistance will also tend to accumulate more in the region where that applies(where the resistivity is greater). This is a phenomenon which is typically observed, but in a different case, as "Skin effects":http://en.wikipedia.org/wiki/Skin_effect" [Broken]
    Be sure however to note that the most valuable equation for conserving the physical aspect of this problem, the "continuity equation" is left unchanged:
    [tex] \overrightarrow{\nabla} \cdot \overrightarrow{j} = -\frac{\partial \rho}{\partial t} [/tex] which ensures the integrity of the solution.
    I pray that shed some light on the issue at hand,
    Daniel
     
    Last edited by a moderator: May 5, 2017
  10. Sep 10, 2011 #9
    I love physics

    thank u very much Daniel
     
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