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Electromagnetic Fields and Energy - Chapter 5 ... - MIT

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Sec. 5.10 Three Solutions 49<br />

The integral can be carried out for any given distribution of potential. In this particular<br />

situation, the potential of the surface at y = b is uniform. Thus, integration<br />

gives<br />

<br />

16v 1<br />

A<br />

for m <strong>and</strong> n both odd<br />

mn = mnπ 2 sinh(k mn b) (15)<br />

0 for either m or n even<br />

The desired potential, satisfying the boundary conditions on all six surfaces, is given<br />

by (10) <strong>and</strong> (15). Note that the first term in the solution we have found is not<br />

the same as the first term in the two­dimensional field representation, (5.5.9). No<br />

matter what the ratio of a to w, the first term in the three­dimensional solution has<br />

a sinusoidal dependence on z, while the two­dimensional one has no dependence on<br />

z.<br />

For the capacitive attenuator of Fig. 5.5.5, what output signal is predicted by<br />

this three dimensional representation? From (10) <strong>and</strong> (15), the charge on the output<br />

electrode is<br />

where<br />

a<br />

w ∂Φ<br />

q = − o<br />

0 0<br />

dxdz ≡ −C M v (16)<br />

∂y y=0<br />

∞ ∞<br />

64 <br />

C M = oaw<br />

k mn<br />

π 4<br />

m 2 n 2 sinh(k mnb)<br />

m=1 n=1<br />

odd odd<br />

With v = V sin ωt, we find that v o = V o cos ωt where<br />

V o = RC nωV (17)<br />

Using (16), it follows that the amplitude of the output voltage is<br />

∞ ∞<br />

V o<br />

ak mn<br />

=<br />

<br />

U 2πm 2 n 2 b<br />

sinh<br />

m=1 n=1 (kmn a) a<br />

odd odd<br />

(18)<br />

where the voltage is normalized to<br />

<strong>and</strong><br />

U =<br />

128 o wRωV<br />

π 3<br />

k mna =<br />

<br />

(nπ)<br />

2<br />

+ (mπ) 2 (a/w) 2<br />

This expression can be used to replace the plot of Fig. 5.5.5. Here we compare the<br />

two­dimensional <strong>and</strong> three­dimensional predictions of output voltage by considering<br />

(18) in the limit where b a. In this limit, the hyperbolic sine is dominated by one<br />

of its exponentials, <strong>and</strong> the first term in the series gives<br />

b<br />

ln V o<br />

→ ln 1 + (a/w)<br />

U 2 − π 1 + (a/w) 2 (19)<br />

a<br />

In the limit a/w 1, the dependence on spacing between input <strong>and</strong> output electrodes<br />

expressed by the right h<strong>and</strong> side becomes identical to that for the twodimensional<br />

model, (5.5.15). However, U = (8/π 2 )U regardless of a/w.<br />

This three­dimensional Cartesian coordinate example illustrates how the orthogonality<br />

property of the product solution is exploited to provide a potential

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