Fields and Waves I Spring 2005 Homework 4. Due 8 March 2005

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Melvyn Townsend
5 years ago
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1 Homewok 4 Due 8 Mach 005. Inceasing the Beakdown Voltage: This fist question is a mini design poject. You fist step is to find a commecial cable (coaxial o two wie line) fo which you have the following infomation: the capacitance pe mete, the type of mateial used fo the insulato (to make it simple to gade this poblem, choose a cable that uses solid polyethylene (PE) fo the insulato), the dimensions of the inne and oute conductos, and the maximum electic field o voltage that the cable can sustain. The latte paamete may not be given, but since you will know the insulating mateial, you can look up the value of its beakdown field. a. Using the paametes you have fo you selected cable, calculate the capacitance fom fist pinciples. That is, find D, E, and V as functions of adius in the egion between the two conductos and then use this infomation to detemine the capacitance, eithe using the chage o the enegy method. If you use the enegy method it is not necessay to find the voltage. This calculation was done in the lectues. Thus, it is only necessay to epoduce it hee and put in the coect vaiables.

2 Homewok 4 o Fictitious Belden 75 Ohm Cable RG59 Inne wie diamete 0.03 inches Oute wie diamete 0.44 inches Insulato is Foam Polyethylene with anε that poduces a velocity of popagation of 83% the speed of light. Beakdown field up to 50kV/mm Fo this cable with PE insulation. a = 08. mm, b = 366. mm, ε = = 44. and the (. 64) capacitance is c = 90 pf pf m = 7. 5 ft

3 Homewok 4 We wish to impove on this cable in some way. Ou specific goal will be to change the insulato so that the beakdown voltage inceases while maintaining the same capacitance pe unit length. We will assume that the inne and oute conductos emain the same. We will only change the insulato. The appoach we will take will be to eplace the PE with two o thee othe insulatos as shown. The cente insulato will be chosen to have the lagest beakdown voltage, the othe one o two insulatos will be chosen to obtain the same capacitance. Recall that each will have a diffeent dielectic constant. To find mateial popeties, you can go to MatWeb which has almost any popety one can think of fo commecially available mateials. b. Begin by solving fo the capacitance of a coaxial cable with insulating egions. Assume that the fist insulato is in the egion a b and the second insulato is in the egion b c whee a is the adius of the inne conducto and c is the adius of the oute conducto. The dielectic constant of egion is ε and the dielectic constant of egion is ε. Again, do this fom fist pinciples and find D, E, and V as functions of adius in the egion between the two conductos and then use this infomation to detemine the capacitance, eithe using the chage o the enegy method. If you use the enegy method, it is not necessay to find the voltage. The good news is that D is the same because the dielectic popeties do not show up in Gauss Law. Thus D a = ρ a $ s so that E s a = ρ a $ with a diffeent epsilon in each ε eation. To find the voltage we need to integate the electic field. Fo the oute insulato ρs a V() = E dl = a$ ( a$ d + a$ d + a$ dz c φ φ z ) while fo the inne insulato c ε ρs a V() V() b = E dl = a$ ( a$ d + a$ d + a$ dz) b φ φ z thus we have that b ε s a V d sa c () = ρ ρ = ln fo the oute insulato and c ε ε ρ a c s ρs a V b d ρsa b () ln = = ln. The total voltage diffeence between the oute ε b ε ε 3

4 Homewok 4 ρsa c ρsa b c b and inne conducto is then V( a) = ln + ln = ρsa ln + ln. The ε b ε a ε b ε a capacitance pe unit length is then Q ρs πa πεε C = = =. Note that this c b c b c b ρsa ln + ln ρsa ln + ln εln + ε ln ε b ε a ε b ε a b a could also be detemined by taking the paallel combination of the two capacitos fomed by the two insulatos. C = πε and C = πε. Since the poblem stated that fist c b ln ln b a pinciples must be used, this latte appoach cannot be used to solve the poblem. Howeve, it can be used to check the answe. To get a highe value fo the opeating voltage, we can choose an insulato fom the MatWed site. Thee is a popety seach page that makes this vey simple. On this page, select a value fo the Dielectic Stength of at least 00 and maybe a dielectic constant of.45 (about) to see if anything fits the bill. Unfotunately, nothing shows up. Thus, just pick a dielectic constant that is close. The lowest dielectic constant will be the easiest to match. Thus, choose Saint-Gobain Noton FEP Fluoopolyme Film which has a dielectic constant of. and a dielectic stength of 36. We also need something with a highe dielectic constant to make up fo the lowe value with this mateial. We can choose many diffeent mateials. Ty DuPont Kapton 0FN66 Polyimide/FEP Composite Film with 3. and 7. Then use the lowe value dielectic constant on the inne egion and the highe in the oute egion. Then if we choose the thickness of the inne egion to be 0.075inches o.9mm, we get the same value of capacitance using two mateials with lage beakdown fields. 4

5 Homewok 4 To complete you design, you need to identify two new dielectic mateials, one of which has a highe beakdown voltage than PE and the othe has a dielectic constant that compensates fo the fist dielectic in the capacitance. You can choose the distance b to get the capacitance you need. If you cannot make two insulatos wok, you will have to ty thee. c. What ae the final paametes of you design? Demonstate that the capacitance is the same and the beakdown voltage is lage. The final paametes ae theefoe the Saint-Gobain dielectic in the inne egion and the Dupont in the oute egion. The maximum field occus in the inne egion so the dielectic stength is limited by the popeties of the Saint-Gobain mateial. Note that using two diffeent dielectics actually does not give the full impovement in this case because the field in the lowe dielectic egion is inceased elative to the unifom case. Hee, fo example, the field at =a is given by the usual expession fo E s a = ρ a $. ε Since the chage density must be the same if the capacitance is the same, the field is lage hee by the atio ε ε = 44.. =6.. Fotunately the dielectic stength inceases by moe than this value 36 = 57. so we eally have made things bette. Any student who 50 ecognizes this gets exta cedit. 5

6 Homewok 4. Using a Speadsheet to Find Capacitance: Assume that you have the following two dimensional configuation of conductos (all dielectics ae ai). Fo this poblem, see the Excel speadsheets. 6

7 Homewok 4 Assume that the voltage on conducto is -00V and on conducto is +00V. The diamete of each is 0mm, the oute box is 50mm by 50mm, and the distance between the conductos is also 0mm. a. Use the speadsheet method to find the capacitance pe unit length of this shielded, two wie tansmission line. b. Poduce a plot showing at least 8 equipotentials. Sketch a epesentative set of electic field lines on this plot. c. Find the chage pe unit length. d. Using a standad analytic fomula, find the capacitance pe unit length of the two wie line without the shield. e. Assume that about half of the lowe half of the insulating egion is filled with a dielectic withε = 8ε o. Repeat all steps and find the capacitance pe unit length. 7

8 Homewok 4 3. Poisson s Equation: The electic scala potential in a spheical egion 0 a is given by V () = V o. This egion is known to be filled with chage with an unknown density distibution ρ = ρ(). Detemine the chage distibution esponsible fo this potential. This is simply found by applying the Laplacian to this voltage. ρ = V = = = ( 0) = 0 ε V() Vo () Thus, the chage density is zeo o thee is no chage. Any time the voltage is a constant, the chage must be zeo, egadless of the coodinate system. 8

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