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Please Help: Guass's Law and surface integrals

  1. May 7, 2008 #1
    I'm in need of help. I need the formulas for total-electric-flux enclosed for a final exam *tommorow. My teacher (nice guy but total slacker) never did any handouts, and I am not the quickest to catch on when he tried to explain in about 20 minutes the concept of surface integrals. I was just out of calc II and had absolutely no experience with cylindrical or spherical coordinates.
    Spherical surface:
    flux enclosed =∫[0,2π]∫[0,π](sin0)r^2

    infinite plane
    flux enclosed =∫[]∫[]

    Long thin wire
    flux enclosed =∫[0,c]∫[0,2π]

    I am the only student in calc. physics mixed with a group of non-calc students. Surface intergrals did not come up in calc III until about one week ago, and now I'm under pressure. The physics teacher made it clear 'he would make it easy on me', but I just needed to be able to calculate flux enclosed with E and dA given in the cases of:
    1) infinite flat sheet
    2) point charge/ spherical surface
    3) long thin wire

    I just need to be able to set up the integrals. But I can not find the formulas. I'm in desperate need of just the integrals without a lot of extranious information. I tried at the top to fill in what I could.
  2. jcsd
  3. May 7, 2008 #2
    you're going to need the metrics of sphereical and cylindrical coordinates.

    sphere: dr, r*dtheta, r*sin(theta)*dphi

    cylindrical: dr, r*dtheta, dz
  4. May 8, 2008 #3
    I am not sure about the infinite plane case, but for the others, the surface integral is really easy. Becuase these are what are known as primary surfaces (cylinder in cylindrical coordinates, sphere is sphereical and a box in cartesian) you will probably not even have to integrate. For every Physics II book I have looked at, it is introduced as a surface integral but is then done more easily.

    Recall a surface integral can be thought of as a measure of flux, or the amount of a field going into the surface (perpendicularly). For electric fields, in spherical, find the radial component (probably all you have in Physics II), then multiply it by the area of the sphere. For just a sphere when there is only a radial component,

    [tex]\iint_R\vec{E}\cdot\hat{n}dS=\int_{\theta=0}^{2\pi}\int_{\phi=0}^{\pi} E*R^2 \sin \phi d\phi d\theta =E*A=4\pi R^2 E[/tex]

    It is the same logic for the wire only now you have what is going through the side of the cylinder in addition to the top and bottom. Usually, the top in the bottom can be either ignored by symmetry or because it is an infinite wire.

    I hope this helps
    Last edited: May 8, 2008
  5. May 8, 2008 #4
    I realized I should have continued. In these problems, you generally do not want to have the parameters of the surface in your answer. This makes sense becuase the charge inside shouldn't depend on the radius of your arbitrary surface. Notice in the euqation i put in my above post, the R cancels. Recall that usually, [tex]E\propto\frac{1}{r^2}[/tex], which when r is not changing, is just [tex]E\propto\frac{1}{R^2}[/tex].

    For a wire, you will usually have the total charge. Therefore the charge you are using for the E-field is [tex]\frac{q}{L}[/tex] and the area of the cylinder is [tex]2\pi RL[/tex] so the L cancels.
  6. May 8, 2008 #5
    Sorry, one more thing. Note that in physics there is not universal notation for the spherical system. The poster above me called the angle from the Z-axis [tex]\theta[/tex] while I called it [tex]\phi[/tex]. It is different in every book. In Shey's Div, Grad, Curl and All That, it is one way while in Sandri Hassani's Mathematical Methods for Students of Physics and Related Field, it is the other way.
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