# hw4 - ECE3025 Problem Set 4 Solutions r 6.10 Let S = 100...

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ECE3025 Problem Set 4 Solutions 6.10. Let S = 100 mm 2 , d = 3 mm, and ± r = 12 for a parallel-plate capacitor. a) Calculate the capacitance: C = ± r ± 0 A d = 12 ± 0 (100 × 10 6 ) 3 × 10 3 =0 . 4 ± 0 =3 . 54 pf b) After connecting a 6 V battery across the capacitor, calculate E , D , Q , and the total stored electrostatic energy: First, E = V 0 /d =6 / (3 × 10 3 ) = 2000 V / m , then D = ± r ± 0 E =2 . 4 × 10 4 ± 0 . 21 μ C / m 2 The charge in this case is Q = D · n | s = DA . 21 × (100 × 10 6 )=0 . 21 × 10 4 μ C=21pC Finally, W e =(1 / 2) QV 0 . 5(21)(6) = 63 pJ . c) With the source still connected, the dielectric is carefully withdrawn from between the plates. With the dielectric gone, re-calculate E , D , Q , and the energy stored in the capacitor. E = V 0 /d / (3 × 10 3 ) = 2000 V / m , as before .D = ± 0 E = 2000 ± 0 =17 . 7nC / m 2 The charge is now Q = . 7 × (100 × 10 6 )nC=1 . 8pC . Finally, W e / 2) QV 0 . 5(1 . 8)(6) = 5 . 4pJ . d) If the charge and energy found in (c) are less than that found in (b) (which you should have discovered), what became of the missing charge and energy? In the absence of friction in removing the dielectric, the charge and energy have returned to the battery.

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hw4 - ECE3025 Problem Set 4 Solutions r 6.10 Let S = 100...

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