(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

For a time independant or static situation

[tex] -\int_{a}^{b} \vec{E} \bullet d\vec{l} = \Phi(b) - \Phi(a) [/tex]

a) Show that for any closed path

[tex] \oint \vec{E} \bullet d\vec{l} = 0 [/tex]

b)Using Stokes Theorem show that

[tex] \oint \vec{E} \bullet d\vec{l} = 0 [/tex] implies [tex] \vec{\nabla} \times \vec{E} = 0 [/tex]

c) [itex] -\int_{a}^{b} \vec{E} \bullet d\vec{l} = \Phi(b) - \Phi(a) [/itex] was derived by integrating [itex] \vec{E} = -\vec{\nabla} \Phi [/itex]. Using Cartesian coordiants show that this implies [itex] \vec{\nabla} \times \vec{E} = 0 [/itex]

2. The attempt at a solution

a) For any closed path the endpoints a and b are the same. So

[tex] -\int_{a}^{a} \vec{E} \bullet d\vec{l} = \Phi(a) - \Phi(a) = 0[/tex]

Is that it? Just one line?? Is there more i should be considering?

b) Stokes theorem is

[tex] \int_{S} \left(\nabla \times v) \bullet da = \oint_{P} v \bullet dl [/tex]

so here we already know that

[tex] \oint \vec{E} \bullet d\vec{l} = 0 [/tex]

thus the right hand side is zero

[tex] \int_{S} \left(\nabla \times E) \bullet da = 0 [/tex]

ok since the integral is zero either one of two things - either the integral was over some closed surface S and/or the integrand was zero

or the curl of E is perpendicular to da

how would i prove thi formally??

c) All i do is simply take the curl on both sides of

[itex] \vec{E} = -\vec{\nabla} \Phi [/itex]

since the curl of a gradient is zero we prove that it is zero

since the proof asks to use Cartesian coords i guess i have to do something like this

[tex] - \nabla \times \left(\frac{\partial \Phi}{\partial x} +\frac{\partial \Phi}{\partial y} +\frac{\partial\Phi}{\partial z} \right) [/tex]

and prove that the above expression is zero

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# Easy EM problem

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