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How do I get E field?, etc.[Griffiths EM example 7.2, p 287.]

  1. Sep 2, 2011 #1
    1. The problem statement, all variables and given/known data

    two long cylinders (radii a and b) are separated by material of conductivity (sigma). if they are maintained at a potential difference V, what current flows from one to the other, in a length L?

    2. Relevant equations

    Please see the attachment

    3. The attempt at a solution

    solution for the problem is already in the book because it is an example problem, but I don't understand how I can get the expression for E.
    I am also confused why I have to use line charge density on the inner cylinder. Shouldn't I use volume charge density since the inner cylinder has volume? or at least surface charge density.


    Thank you
     

    Attached Files:

  2. jcsd
  3. Sep 3, 2011 #2

    ehild

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    Representing the electric field by the field lines, the intensity of the electric field is equal to the number of field lines crossing the unit area.
    You know that 4πε0 electric field lines emerge from a unit positive charge.
    The cylinders have be of metal, otherwise the current from the source could not be maintained. (The arrangement is a cylindrical capacitor). As they are of metal, the charges accumulate on the surfaces of the cylinders. As the cylinders are very long the field outside them can be ignored, and the field in between has cylindrical symmetry: it changes only radially. If the surface charge density (charge of unit surface) is ρs, E=ε0ρs at the surfaces of the cylinders.
    The line charge is the charge at the surface of a piece of cylinder of unit length. How is it related to the surface charge density?
    The line charges are equal on both cylinders, but the surface charge densities are different, and the electric field is different too, it changes along the radius from the inner cylinder to the outer one.
    How is E related to r?

    ehild
     
  4. Sep 3, 2011 #3
    Representing the electric field by the field lines, the intensity of the electric field is equal to the number of field lines crossing the unit area.

    I thought the number of electric field line is arbitrary for a given charge, but does increase/decrease with the charge strength or distance from the charge. Should they be "equal" to the electric field intensity? It would have 1/r dependence in 2d, and 1/r^2 dependence in 3d. Am I right?

    You know that 4πε0 electric field lines emerge from a unit positive charge.

    I honestly don't understand what you meant by this. When I write down the expression for E field when there is a single positive charge, I see 4[itex]\pi[/itex][itex]\epsilon[/itex]0 in the denominator.

    The cylinders have be of metal, otherwise the current from the source could not be maintained. (The arrangement is a cylindrical capacitor). As they are of metal, the charges accumulate on the surfaces of the cylinders. As the cylinders are very long the field outside them can be ignored, and the field in between has cylindrical symmetry: it changes only radially. If the surface charge density (charge of unit surface) is ρs, E=ε0ρs at the surfaces of the cylinders.

    Sorry, I don't understand why E=[itex]\epsilon[/itex]0ρs
    I find in the book that when surface charge density is considered, E=(1/4[itex]\pi\epsilon[/itex]0)*[itex]\int\sigma[/itex]/(r2)[itex]\hat{r}[/itex]da

    The line charge is the charge at the surface of a piece of cylinder of unit length. How is it related to the surface charge density?

    maybe I can think a surface charge density as square of line charge density, but I haven't seen this relationship anywhere

    The line charges are equal on both cylinders, but the surface charge densities are different, and the electric field is different too, it changes along the radius from the inner cylinder to the outer one.

    if the same amount of charges are distributed uniformly on both surfaces of cylinders, the surface charge densities should be the same. However, are the line charge densities equal? I don't understand why E field should be different, because E field will be affected by both cylinders and I have to consider the distance from each cylinder to the point in interest

    How is E related to r?

    I think E has inverse r dependence.



    um... so..... total chaos. I'm really sorry, but I am really having difficulty in this.

    Thank you.
     
  5. Sep 3, 2011 #4

    vela

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    You can use Gauss's law to calculate E between the two conductors, taking advantage of the cylindrical symmetry of the configuration.
     
  6. Sep 3, 2011 #5

    ehild

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    You are right, I made a mess with the 4π and ε0. It was very hot here... And you do not need really what I wanted to explain about electric field lines and surface charges, but it is useful to imagine the problems, so I write it again -hopefully without errors. If you are not interested, go to **

    The electric field of a point charge is
    E=Q/(4πε0r2). According to Gauss' Law, the surface integral of E around a sphere enclosing the charge is Q/ε0.
    When representing the electric field by the field lines, we take that the electric field strength is equal to the number of field lines crossing a unit surface, normal to the field, so the number of electric field lines crossing a total sphere around the point charge is Q/ε0. That is why it is said that Q/ε0 field lines emerge from a charge Q. The field lines originate at the positive charges and end at the negative ones (unless they go to infinity).

    When the charges are distributed on a metal surface we know that the electric field is zero in the metal. So all the field lines go outward from the metal if the surface charge is positive, and inward if it is negative. The potential of a metal surface is the same at any point: it is an equipotential surface. The electric field is the negative gradient of the potential, its direction is normal to the equipotential surface.
    We said that Q/ε0 electric field lines emerge from a charge Q. A unit surface of the metal has ρs (surface) charge, which means ρs0 field lines per unit area. That is, the electric field strength is ρs0 at the metal surface.
    The surface charge density on the inner cylinder (with radius a) multiplied by the area of the cylindrical surface of unit length gives the line charge density, λ. It is the same for both cylinders, but the surface charge densities are different, as the outer surface is larger than the inner one.
    λ=2πaρs, that is, ρs=λ/(2πa), and the electric field at the inner cylinder surface is E=λ/(2πaε0).

    **
    The problem asks the current when the potential difference between the cylinders is given. The potential difference is obtained by integrating the electric field intensity between r1=a and r2=b.
    The electric field at a point r from the centre of the inner cylinder can be obtained by Gauss Law: E 2rπ L = Q/ε0 where Q=λL is the total charge of the cylinder of length L and λ is the line charge density. So E=λ/(2πε0r). There is r in the denominator in this case, instead of r^2 valid for spherical symmetry.

    The line charge density λ is obtained then from the potential difference. The electric field at the surface of the inner cylinder is obtained from the line charge density. The current is obtained from the conductivity σ and the electric field strength i=σE. The total current is equal to the current density multiplied by the area of the inner cylinder.

    ehild
     
    Last edited: Sep 3, 2011
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