Recent content by Bert

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    Rigidbody Dynamics Exercises: Find Online

    can I ever find solved exercises?
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    Rigidbody Dynamics Exercises: Find Online

    Where can I find online a good set of rigid body dynamics exercises? Thanks.
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    Understanding Boundary Conditions: Why 33.22abe is False

    Sorry for my bad English I mean a proof by contradiction.
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    Understanding Boundary Conditions: Why 33.22abe is False

    Oké thanks I understand that. But by contradiction if 33.22b would be discontinue than equation 33.22a would be false why can he say that? The first argument about the continuity of 33.2b is clear but for me not the second contraction proof.
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    Understanding Boundary Conditions: Why 33.22abe is False

    what do mean the author by the red underline line? http://img164.imageshack.us/img164/7404/contfv6.jpg Why would 33.22abe false? Thanks.
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    Continuity of the first Maxwell equation.

    Thanks if I define my y component as the parallel one. Is than the proof oké?
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    Continuity of the first Maxwell equation.

    Suppose that we will proof the continuity of the first maxwell equation: So we have div(\vec{E})=\frac{1}{\epsilon _0} \rho than \iiint \ div(\vec{E}) = \oint_v \vec{E} d\vec{s}=\iiint \frac{1}{\epsilon _0 } \rho than follewed E_{y1} l -E_{y2}l=Q Therefore E must continue is this a...
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    Finding the Electric Field Twice: Solving for Potential

    the first -k \int y^2 dx because the path over y is 0 so y=0 therfore, I think that we get - k \int 0 \ dx and this is constant so - k \int 0 \ dx =c is that wrong?
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    Finding the Electric Field Twice: Solving for Potential

    for the first y=0 so some ct so -k ct for the second \int -k \ 2xy = -k \ \frac{2}{2}xy^2 =-k \ xy^2 for the last one -k \ 2 \ y \ \frac{1}{2} z^2=-k \ y \ z^2 so finaly -k -kxy^2 - kyz^2 is this oké?
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    Finding the Electric Field Twice: Solving for Potential

    oké I need to remove x y components en substitute somthing like y=kx etc ?
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    Finding the Electric Field Twice: Solving for Potential

    sorry I don't see why my path hold not for every arbitrary (x,y,z) I will like to calculate the potential in every point in space in one time?
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    Finding the Electric Field Twice: Solving for Potential

    Thus the problem is path dependent? so i calculate the rot from the eletric field and that gives 0. Thus a conservative field? why not?
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    Finding the Electric Field Twice: Solving for Potential

    Given is the electrical field &\mathbf{E}& = k[y^2 \^{&\mathbf{x}&} + (2xy + z^2)\^{&\mathbf{y}&} + 2yz \^{&\mathbf{z}&}] I will like to find the potential so I integrate and become V(r) = -k (y^2x + xy^2 + z^2y + yz^2) then I try to find the elektrical field again, so I differentiate...
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