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Deriving electrodynamic equations

  1. Oct 10, 2011 #1
    Hey all. I am taking my second college physics course (electromagnetic physics) and am looking for some help deriving the equations. I found it very helpful to know how to derive many of the equations in my first physics course. So far we have studied e fields, guass's law, capacitors, resisters, potential, and power. The equations are begining to pile up. My professor said they can all be derived from Va-Vb=∫E ds

    Is he accurate?
     
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  3. Oct 10, 2011 #2

    Born2bwire

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    They can all pretty much be derived from the Maxwell Equations with sufficient knowledge of vector calculus.
     
  4. Oct 11, 2011 #3
    I am in vector calculus as well, but is it really necessary to derive them that way? Basic calculus is the only prerequisite for the course. Does anyone know how to derive them from ∫E ds
     
  5. Oct 11, 2011 #4

    Matterwave

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    The equation you gave is merely a definition of the voltage, I don't think you can "derive" Gauss's law from that.

    Which equations specifically are you trying to derive?
     
  6. Oct 11, 2011 #5

    Born2bwire

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    That would be vector calculus. Gauss' Law is
    [tex] \nabla \cdot \mathbf{D} = \rho [/tex]
    Taking the integral over volumetric space and using the divergence theorem,
    [tex] \int \mathbf{D} \cdot d \mathbf{S} = \int \rho dV = Q_{enclosed}[/tex]
    If we assume a homogeneous medium then finally,
    [tex] \int \mathbf{E} \cdot d \mathbf{S} = \frac{Q_{enclosed}}{\epsilon} [/tex]


    Now in electrostatics, Maxwell's Equations state that the curl of the electric field is zero. That is,
    [tex] \nabla \times \mathbf{E} = 0[/tex]
    This allows us to represent the electric field as the gradient of a scalar since the curl of a gradient is always zero. Thus, we choose this scalar to be the electric potential.
    [tex] \mathbf{E} = -\nabla V [/tex]
    If we take the line integral of the electric field from some point B to A we get via the gradient theorem,
    [tex] \int_b^a \mathbf{E} \cdot d\mathbf{\ell} = V_b - V_a [/tex]
    which is path independent because the electrostatic field is conservative (by virtue of being curl free). Finally, we can use Gauss' Law to see that
    [tex] \nabla^2 V = -\frac{\rho}{\epsilon} [/tex]
    which is Poisson's Equation.
     
  7. Oct 11, 2011 #6
    Excellent that's helpful! Thank you
     
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