754_Physics ProblemsTechnical Physics

754_Physics ProblemsTechnical Physics - 94 Capacitance and...

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Unformatted text preview: 94 Capacitance and Dielectrics P26.53 (a) Consider a gaussian surface in the form of a cylindrical pillbox with ends of area A ′ << A parallel to the sheet. The side wall of the cylinder passes no flux of electric field since this surface is everywhere parallel to the field. Gauss’s law becomes EA ′ + EA ′ = (b) Q away from the positive and 2 ∈A toward the negative sheet. Together, they create a field of Q . ∈A Assume that the field is in the positive x-direction. Then, the potential of the positive plate relative to the negative plate is z + plate ∆V = − z + plate E ⋅ ds = − − plate (d) directed away from the positive sheet. In the space between the sheets, each creates field E= (c) Q Q A ′ , so E = 2 ∈A ∈A Q Qd i ⋅ − idx = + . ∈A ∈A − plate e Capacitance is defined by: C = j Q Q ∈ A κ ∈0 A = = = . d d ∆V Qd ∈ A Additional Problems P26.54 (a) C= (c) Q ac LM 1 + 1 OP + LM 1 + 1 OP = 3.33 µF N 3.00 6.00 Q N 2.00 4.00 Q = C b ∆V g = b 2.00 µFga90.0 V f = 180 µC −1 ac −1 ac Therefore, Q3 = Q6 = 180 µC ib d ga f Q df = C df ∆Vdf = 1.33 µF 90.0 V = 120 µC (b) (d) Q3 C3 Q ∆V6 = 6 C6 Q ∆V2 = 2 C2 Q ∆V4 = 4 C4 ∆V3 = UT = 180 µC = 3.00 µF 180 µC = = 6.00 µF 120 µC = = 2.00 µF 120 µC = = 4.00 µF = af 1 C eq ∆V 2 2 = 60.0 V 30.0 V 60.0 V 30.0 V ja 1 3.33 × 10 −6 90.0 V 2 e f 2 = 13.4 mJ FIG. P26.54 ...
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