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Electrodynamic doubt

  1. Oct 4, 2014 #1
    Hello, I have a doubtaboutthe use of Gauss theorem in the presence of a coductor.

    If I have a infinite cylindric long shell with a charge density ƍ of radius b and inside of it a cylindrical conductor of radius a, also of infinite lenght, what will be the electric field in a point located between a and b.

    I think it will be zero because there is no charge into the shell, but I am nt 100% sure
  2. jcsd
  3. Oct 4, 2014 #2
    If the conducting cell is infinitely long, than you're right.
    This is like a Faraday cage.
  4. Oct 4, 2014 #3
    Thanks for your anwer, I thought that some charge might be in the surface of the conductor; then I gess the potential will also be a constant inside of the radius b.
  5. Oct 4, 2014 #4


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    Staff: Mentor

    Wait a minute! Are we talking about the region inside an infinite cylinder of constant charge density but outside a smaller concentric conductor?
  6. Oct 4, 2014 #5
    I think we do.
  7. Oct 5, 2014 #6
    No, I might have said it wrong, there is a cylinder of radius a and infinite lenght wich is a conductor and it is uncharged. there also is a cylindric shell of radius b; this one has a density of charge RO.
  8. Oct 5, 2014 #7
    Give us the sketch Einstein. I don't trust you o0)
  9. Oct 5, 2014 #8
    Ejemplo de teoría (evaluación continua)
    Se tiene un cilindro conductor descargado de radio a y longitud infinita. Dicho
    conductor está rodeado de una cáscara cilíndrica de radio b y densidad de carga
    uniforme .
    1) Calculad el campo electrostático en todos los puntos del espacio.
    2) Calculad el potencial electrostático en todos los puntos del espacio
    3) Analizad el comportamiento del campo y del potencial a grandes distancia y discutid
    su significado.

    This is the sketch; the best translation I can provide is:
    Given an uncharged conductor cylinder of radius a and infinite lenght, said conductor is surrounded by a cylindric shell of radius b and uniform density of charge RO.
    Calculate the electrostatic field in every point of space
    Calculate the electrostatic totential in all points of space
    Analyze the behaviour of the field and potential at big distances and discuss its meaning
  10. Oct 5, 2014 #9
    Ok then. Everywhere inside the cylindric shell E-field =0 (as said before)
  11. Oct 5, 2014 #10
    Thanks again for all your help, now I only have to find a suitable point to put the orygin of potential.
  12. Oct 5, 2014 #11


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    Gold Member
    2017 Award

    Just put your origin somewhere on the cylinder's axis, which we choose as the [itex]z[/itex] axis. The problem is translation invariant along this axis and under rotations around this axis. So you introduce cylinder coordinates [itex](r,\varphi,z[/itex]. I guess that the whole question is about electrostatics. Then the electric field has a potential [itex]\vec{E}=-\vec{\nabla} \Phi[/itex], and due to the symmtries this potential depends only on [itex]r[/itex].

    Then Gauß's Law tells you
    [tex]\vec{\nabla} \cdot \vec{E}=-\Delta \Phi=\rho. \qquad (1)[/tex]
    The charge density is
    [tex]\rho=\sigma \delta(\rho-b).[/tex]
    The Poisson equation (1) reads for our case
    [tex]\frac{1}{r} \frac{\mathrm{d}}{\mathrm{d} r} \left (r \frac{\mathrm{d} \Phi}{\mathrm{d} r} \right )=-\rho.[/tex]
    Now we have nearly everywhere [itex]\rho=0[/itex]. So we solve the differential equation for [itex]\rho=0[/itex]. We simply can multiply by [itex]r[/itex] and integrate ones. This gives
    [tex]r \frac{\mathrm{d} \Phi}{\mathrm{d} r}=C_1.[/tex]
    Dividing by [itex]r[/itex], and integrating again gives
    [tex]\Phi(r)=C_1 \ln \left (\frac{\rho}{\rho_0} \right ). \qquad(2)[/tex]
    Here, I have written the second integration constant such as to make the log's argument dimensionless as it must be.

    Now we need to work out the boundary conditions. First of all in the conducting inner cylinder we must have [itex]\vec{E}=0[/itex] and thus [itex]\Phi=\text{const}.[/itex] For convenience we set [itex]\Phi(r)=0[/itex] there. Now there is no charge on this inner cylinder, and the potential must be smooth on the boundary [itex]r=a[/itex]. Thus we have
    [tex]\Phi(r)=0 \quad \text{for} \quad r \leq b.[/tex]
    Now we must work in the surface-charge density at [itex]r=b[/itex], where according to (1) the potential must be continuous but its first derivative must jump such that the [itex]\delta[/itex] distributions singularity is correctly described. Multiplying (1) with [itex]r[/itex] and integrating wrt. [itex]r[/itex] between [itex]r=b-0^+[/itex] and [itex]r=b+0^+[/itex] yields
    [tex]b \left [\Phi'(b+0^+)-\Phi'(b-0^+) \right ]=-\sigma b.[/tex]
    Since [itex]\Phi'(b-0^+)=0[/itex], we must have
    Taking the derivative of our solution (2), we find
    [tex]\frac{C_1}{b}=-\sigma \; \Rightarrow \; C_1=-\sigma b.[/tex]
    The constant [itex]\rho_0[/itex] must be chosen such that [itex]\Phi(b)=0[/itex], i.e., [itex]\rho_0=b[/itex]. Thus our solution for the potential reads
    0 & \text{for} \quad r \leq b, \\
    -\sigma b \ln \left (\frac{r}{b} \right ) & \text{for} \quad r>b.
    The electric field is
    [tex]\vec{E}=-\vec{e}_r \Phi'(r)=\begin{cases}
    0 & \quad \text{for} \quad r<b \\
    \sigma b/r \vec{e}_r & \quad \text{for} \quad r>b.
    Note, that the normal component of [itex]\vec{E}[/itex] which is [itex]E_r[/itex] makes the correct jump at [itex]r=b[/itex], namely the surface-charge sensity [itex]\sigma[/itex]. So we have found the complete solution of the problem.
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