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Homework Help: The integral

  1. Sep 25, 2009 #1
    1. The problem statement, all variables and given/known data

    I am soooo lost. I don't even know if this is the right forum...... But where is the bridge between Calculus and Physics? I can Integrate equations, but when it comes to physics, i for one, don't know when to integrate; two, i don't see how you find the constants to remove from the integral; and three, Even given the integral formula for an equation, i still don't know what im doing. ?? finding the electric field of an object?? i thought Electric field was (1/4pi*epsilon naught)(Q/r^2). So how do i find the E field for different shapes?

    another example...

    ex. Va-Vb=SE.dl

    E-Electric Field
    dl-small segments of length

    I don't know how to use the equation;

    Or, electric flux,

    Flux=SE . dA

    What am i not understanding. Please help

    2. Relevant equations

    SE.dA - flux
    SE.dL - potential difference
    SE.dr - is = SE.dL when im dealing with a charge line within a cylinder(Gaussian surface)

    3. The attempt at a solution
  2. jcsd
  3. Sep 25, 2009 #2
    There are two things that are difficult for us students dealing with integral calculus in physics.

    The easy part is solving the integral (You usually don't end up with things that are too complicated, and if you do, chances are you're allowed to use a look-up-table).

    The tricky bit is constructing the integral and understanding what it means, and the even trickier bit is looking at it, and finding special cases where you can just do it in your head without mucking up the math in between.

    The Gauss' Law example you've brought up is perfect to demonstrate this point.

    Consider a uniform electric field [tex]\vec E[/tex] through a flat surface, with area [tex]A[/tex], which we will define as a vector whose magnitude is the area of the surface, and whose direction is perpendicular to the surface:
    [tex]\vec A\equiv |A|\hat n[/tex]

    Now we will define the flux of the electric field through this area:
    It is the component of the electric field, in the direction perpendicular to the surface.
    Or, in vector notation, if that doesn't frighten you:
    [tex]\Phi = \vec E\cdot \vec A[/tex]

    You can see from this definition, that if the direction of the field is perpendicular to the surface, then the flux is just: [tex]\Phi=EA[/tex]

    Now, since the flux is a scalar, we can say that it is additive. So if we have two flat surfaces, each with a field going through it, [tex]\vec E_1, \vec E_2, \vec A_1, \vec A_2[/tex]
    Then the total flux, is the sum of the flux through each of the surfaces:
    [tex]\Phi = \vec E_1\cdot \vec A_1 + \vec E_1\cdot \vec A_1[/tex]

    Taking this definition further, we get the integral definition of flux through a Gaussian surface:
    This definition means:
    The total flux, is what we get if we take a surface, and go tile by tile, and see what part of the field is perpendicular to that tile.

    [tex]\Phi=\int \vec E\cdot \vec dA=\int |E||dA|\cos{\theta}[/tex]

    The interesting special case is when the field is uniform, and always perpendicular to the surface you've chosen, you get that it all narrows down into:


    You hardly ever have to use the integral definition, and this is the one you will usually use.
    Always look for cases where symmetry can get you a uniform field that's always perpendicular to your chosen surface.
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