- #1
stunner5000pt
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- 2
Homework Statement
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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