KAEV IAEW Übung 1 Lösung PDF

Preview

Summary

KAEV Übung...

Description

~ d~s =  @ =  @ E @t @t h!0



A

 ~ dA ~ ⇡ 0, B

~ d~s = |E ~ 1t | · l + |E ~ 2t | · l = 0 E ~ 1t | = |E ~ 2t | |E

)

~ dA ~= D

⇢dV = 0,

h!0

V

~ dA ~ = |D ~ 1n | · l · b + |D ~ 2n | · l · b = 0 D h!0)

h·b

~ 1n | · l · b + |D ~ 2n | · l · b = 0 |D ~ 1n | · l · b = | D ~ 2n | · l · b |D

!

h·l

E0 kV Et1 = Et0 = E0 · cos(60) = = 10 cm 2 ✓ ◆ kV En0 = E0 · sin(60) = 17, 33 cm Dn0 = Dn1 ) "0 · "r0 · En0 = "0 · "r1 · En1 "r0 En1 = En0 · En0 = E0 · sin(60) "r1 p kV " 3 1 r0 · = 4, 33 = E0 · sin(60) · = E0 · 2 4 "r1 cm s✓ ◆2 ✓ ◆2 q kV kV kV 2 2 = + 4, 33 10 |E1| = Et1 = 10, 9 + En1 cm cm cm Et1 ) Et1 = 10 kV cm

kV cm Dn1 = Dn2 ) "0 · "r1 · En1 = "0 · "r2 · En2 "r1 kV 4 · En2 = En1 · = 4, 33 cm 2 "r2 kV ) En2 = 8, 66 cm r q kV kV kV 2 2 = 8, 66 |E2| = En2 = 13, 23 + Et2 + 10 cm cm cm Et2 = Et1 = 10



E~2 10 kV Et0 Et2 cm = 0, 76 = = sin() = E2 E2 13, 23 kV cm



)  = 49, 1

~2 E  = 180  45   = 180  45  49, 1 = 85, 9 Es2 = E2 · cos() = 13, 23

kV kV · cos(85, 9) = 0, 95 cm cm

kV 10 cm Et1 E?s1 = 2, 3 = = En1 4, 33 kV Es1 cm ) E?s1 = 2, 3 · Es1

E?s1,Durchschlag = 10·Es1,Durchschlag !

kV Es1 = En1  20 cm

! Es2 = 0, 95

kV cm

E?s2 = E2 · sin() = 13, 23 · sin(85, 9)

kV kV = 13, 2 cm cm

E?s2 13, 2 kV cm = = 13, 9 kV Es2 0, 95 cm ) E?s2 = 13, 9 · Es2 !

!

E?s2  200 kV cm

Es1 = 4, 33 kV cm

E0 = 20 kV cm ) Es1 = 0, 22 · E0

kV Es1  20 cm !

kV E?s2 = 13, 2 cm !

E?s2  200

)

) E01max

20 kV kV cm = = 91 0, 22 cm

E0 = 20 kV cm ) E?s2 = 0, 66 · E0 kV cm

) E02max =

E0max = 91 kV cm

200 kV kV cm = 303 cm 0, 66...