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Capacitor Fringe Field Strength on Axis

  • Thread starter Opus_723
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  • #1
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Homework Statement



Show that the transformation

[itex]w = e^{z} + z[/itex]

maps the infinite lines [itex]y = \pm\pi[/itex] into semi-infinite lines [itex]u \leq u_{0}, v = \pm\pi[/itex]. This is equivalent to transforming an infinite or edgeless parallel-plate capacitor (z-plane) into a parallel plate capacitor (w-plane). Sketch the equipotentials in the w-plane near the edge of the capacitor plates.

Find the electric field at the plane midway between the two plates (at v = 0) as a function of u.

Homework Equations



I found parametric equations for the equipotentials:

[itex] u = x + cos(\frac{\pi*V}{V_{0}})*e^{x}[/itex]
[itex] v = \frac{\pi*V}{V_{0}} + sin(\frac{\pi*V}{V_{0}})*e^{x}[/itex]


The Attempt at a Solution



But I don't know how to find the electric field as a function of u along the axis. If I had an expression for V I would simply take the gradient, but buried as it is inside these parametric equations, I don't know how to get at it. I managed to elimnate x:

[itex] x = ln(\frac{v-\pi*V/V_{0}}{sin(\frac{\pi*V}{V_{0}})})[/itex]

But I'm not sure how to proceed from there.
 

Answers and Replies

  • #2
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Any help would be greatly appreciated. I'm stuck. However, I'm convinced that my parametric equations are correct, so its just a matter of how to go to the electric field from there. Since V is constant on the u-plane, I only need to take the partial derivative of V with respect to the vertical coordinate v (unfortunate choice of letters, I should have changed that). But I don't know how to take a partial derivative when the potential V is buried inside these parametric equations.
 

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