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Homework Help: Electric and Magnetic Fields

  1. Nov 5, 2013 #1
    1. The problem statement, all variables and given/known data

    Consider a region of space containing static free charge density ρ(r) and current density j(r). What two physical laws determine the electric and magnetic field in this region?

    State the integral form of each law, showing the explicit dependence of each on ρ(r) and j(r)

    2. Relevant equations

    Maxwell's equations in general differential form:

    Maxwell's equations in matter in terms of free charges and currents:

    3. The attempt at a solution

    I think the question is referring to Maxwell's equations, is that right? If so I'm not sure which set of equations I must to be looking at. By "static charge", does the question imply that there are no alternating currents? If that's the case the last term of the last equations must vanish.

    The 2nd set is for inside of materials that are subject to electric and magnetic polarization and I'm not sure if that's the correct assumption here. But if we use this #1 and #4 make explicit dependence on ρ and j, are these the right equations? :confused:

    So the integral forms are:

    ##\oint_S D. da = \int_V \rho_f dV = Q_{fenc}##

    ##\oint H . dl = \int_V j(r) dV = I_{fenc}##

    I'm not sure if this is what the question is asking. So any helps is appreciated.
  2. jcsd
  3. Nov 5, 2013 #2


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    "Static free charge density" means that the free charge density is static; does not change with time. The other interpretation would be for a laundry soap commercial ...

    So there is a charge density everywhere, and also currents (which are flows of charge) ... of course either of these might be zero in some places, but the functions are defined everywhere. You are not given any information about the nature of the currents ... they need not be AC or DC, nor even confined to wires.

    "Consider a region of space" means you are in a vacuum situation ... and since you are not given any materials or material properties, you only have enough information for vacuum solutions: free fields.

    With this information the equations to start with are the 1st and 4th Maxwell equations as given in your first set - except that they asked for the integral statements.

    They did not ask you to simplify or solve the equations!
  4. Nov 5, 2013 #3
    Thank you very much for the clarification. However for writing the 4th equation in integral form, I think the variable r in j(r) represents the linear distance from an origin. So should I somehow change "r" into "a" in order to write down the integral?

    This is what I have so far:

    ##\oint_C B . dl = \mu_0 \int j(r) da + \epsilon_0 \mu_0 \frac{d}{dt} \int E.da##

    The ∫j(r)da is the total current passing through the surface.
  5. Nov 5, 2013 #4


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    r is the radius vector ... it sweeps through all of space. The ".da" is the dot product of the vector with an area element of the surface with boundary curve C; this is the same for both integrals on the RHS. On the LHS it is a line integral along the actual curve C [from the Stokes theorem].

    The integral forms (and explanations!) for Maxwell's equations in SI units can be found here:

    You need not show the integral for the current: the integral form is expressed in terms of the total current flowing through that surface from the LHS; if you do show it you need ".da".

    The final term should be a partial derivative.
  6. Nov 6, 2013 #5
    Thank you for the link. Here is my last question, the link shows how to get from the differential form of Ampere-Maxwell equation to its integral form. But to go in the opposite direction I've applied Stoke's theorem to the LHS of

    ##\oint_C B . dl = \mu_0 I + \mu_0 \epsilon_0 \frac{\partial}{\partial t} \int E . da##

    to get

    ##\int_S (\nabla \times B) . da = \mu_0 \int_S j .da + \mu_0 \epsilon_0 \frac{\partial}{\partial t} \int E . da##

    Now is it correct to just cross out all the integral signs (since this is true for any surface)?
  7. Nov 6, 2013 #6


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    Yes ... because the surface is arbitrary, so make it smaller and smaller ... in the limit you have just the integrands.
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