hw6_solutions

# hw6_solutions - 108 THE STATIC ELECTRIC FIELD &gt;From...

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108 THE STATIC ELECTRIC FIELD >From Example 4-15, the maximum electric potential difference (i.e., voltage rating) between the cylindrical conductors can be written as Ps a b b ICI>balmax = ~ In - = aEmax In - = 2000 V f a a Also from Example 4-27, the capacitance of a coaxial capacitor is c = 21C'd In(bja) (a) For oil with fr = 2.3 and EBR = 15 MV-m- 1 , using the voltage rating specified with a = 5 mmyields In ~ = ICI>bai max = 2000 ~ 0.267 a aEmax (5 x 10- 3 )(0.1)(15 x 10 6 ) from which the maximum capacitance ofthe coaxial capacitor with I = 3 cm can be calculated as G = 21C' x 2.3 x 8.85 x 10- 12 x 3 X 10- 2 ~ 144 F max 0.267 . P (b) For mica with fr = 5.4 and EBR = 200 MV-m- l , repeating the same calculations yield Cmax ~ 451 pF. As observed, using mica allows one to design a capacitor with a much higher value due to the fact that mica has a much higher breakdown electric field (by a factor of'" 13.3) and a higher dielectric constant (by a factor of '"2.35) compared with oil. 4·33. Coaxial capacitor with two dielectrics. (a) This problem can be solved in two different ways. In the first method, we can take advantage of the formula for capacitance per unit length of the coaxial line, as determined in Example 4-27, namely 21C'f C eoax = In(bja) where f is the permittivity of the dielectric filling. TIle configuration shown in Figure 4.70 can be considered to be two 'half-coaxial' capacitors of length I connected in parallel. with capacitances respectively of C - 1 21C' f l r l and C2 = ! 21C'f2rl 1 - 2ln(bja) 2In(bja) which combine additively (capacitors in parallel) to give the total capacitance per unit length of C = 1C' f O(flr + f2r)1 In(bja) . The same result can also be obtained using Gauss's law and noting that the electric field has to be

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## hw6_solutions - 108 THE STATIC ELECTRIC FIELD &gt;From...

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