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Electrodynamics and electrical fields

  1. Apr 23, 2008 #1
    [SOLVED] Electrodynamics and electrical fields

    The question is: A long cylinder carries a charge density that is proportional to the distance from the axis ρ =kr, where k is a constant, r is the distance from the axis. Find electric field inside the cylinder.


    My attempt: Ok, first of all we know from Gauss' law that [tex]\oint_S {{\bf{E}} \cdot d{\bf{a}}} = \frac{1}{{\varepsilon _0 }}Q[/tex].

    We also know that [tex]Q = \int {\rho dV = } \int {k \cdot s \cdot s \cdot dsd\phi dz = \frac{2}{3}\pi kls^3 }[/tex]. Here I have used a Gaussian cylinder of length l and radius s.

    Now I must find [tex]\oint_S {{\bf{E}} \cdot d{\bf{a}}} [/tex].

    First question: The electric field E is given by [tex]{\bf{E}} = \frac{1}{{4\pi \varepsilon _0 }}\frac{q}{{r^2 }}{\bf{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}}
    \over r} }}[/tex]. Is the unit vector r in this case the spherical radius-vector?

    For this cylinder, I write the parametric and differentiate it with respect to phi and z. Then I find the normal-vector as the cross-products of these two.

    Question two: When I find the element da as above, what do I do with E, which is in spherical coordinates?

    I hope you can help me.
     
  2. jcsd
  3. Apr 23, 2008 #2
    Ok, you're making this way to hard for yourself. Assume our cylinder has radius a and height h, and choose our Gaussian surface to be a cylinder of radius r co-axial to the original cylinder. The electric field at any point on the Gaussian surface for a radius [tex]r \leq a[/tex] has constant magnitude and is directed parallel to the surface normal there, so we can simplify the integral to

    [tex]{\int_S E d{A}}[/tex]​

    and then to

    [tex]E{\int_S d{A}}.[/tex]​

    Now, since [tex]{\int_S d{A}}[/tex] is just the surface area of the Gaussian surface, which we know to be [tex]2\pi r h[/tex], we can substitute that in, and we are left with the very simple equation

    [tex]E 2\pi r h = \frac{q_{\mbox{enc}}{{\varepsilon _0 }}.[/tex]​

    I will leave it to you to include the density function in that.
     
    Last edited: Apr 23, 2008
  4. Apr 23, 2008 #3
    Hi foxjwill.

    First, thanks for taking the time to reply. I can see that using the definition of the dot-product simplifies things greatly - but still I would like to know how it can be done "my way". I have just studied vector-calculus most of this week, and I am kinda disappointed in myself that I cannot figure it out.

    The thing that annoys me is particularly the r unit-vector in E. Is this in Cartesian or Spherical coordinates?
     
    Last edited: Apr 23, 2008
  5. Apr 23, 2008 #4
    Whichever is more convenient for the particular problem, as long as the same point is described. I'd suggest actually using cylindrical coordinates.
     
  6. Apr 23, 2008 #5
    Ok, so [tex]
    {\bf{E}} = \frac{1}{{4\pi \varepsilon _0 }}\frac{q}{{r^2 }}\rho}
    [/tex]

    I get that the normal-vector has the rho-component 3*cos(theta). I integrate this over theta and z, which are respectively 0..2Pi and 0..l - and this gives 0, which is not correct.

    Where is my error?
     
    Last edited: Apr 23, 2008
  7. Apr 24, 2008 #6
    I worked it out. d\vec{A} = \vec{n}*dA, and the vectors go out since they are parallel.

    Thanks.
     
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