Lesson_4.5_Homework_and_Solutions

Lesson_4.5_Homework_and_Solutions - Lesson 4.5 Homework...

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Unformatted text preview: Lesson 4.5 Homework Solutions 1. A 100 μF air gap capacitor is charged by a 24-V battery through a 1.2 k Ω resistor. If the area of the capacitor plates is 96 cm 2 , calculate (i) the maximum current into the capacitor plates, (ii) the rate of change of electric field between the plates when the current is maximum, (iii) the magnetic field induced between the plates. Solution: C = 100 μF = 10-4 F, V = 24 V, R = 1200 Ω , A = 96 × 10-4 m 2 (i) The charge on capacitor plates while being charged is given by: Q = Q o (1 – e-t/RC ) Q o = CV = 10-4 F × 24 V = 0.0024 C Q = 0.0024(1 – e-t/RC ) The maximum current is the rate at which the charges reach the plates when t = 0 / 0.0024 t RC dQ I e dt RC- = = / / 3 6 0.0024 0.02 1.2 10 100 10 t RC t RC e e--- = = × × × When t = 0, I o = 0.02 A (ii) The electric field between the capacitor plates is: / o o o Q A Q E A σ ε ε ε = = = 4 2 12 2 1 11 1 1 2 1 0.02 96 10 8.85 10 2.35 10 o o dE dQ I A dt A dt A m C m N m V s ε ε------ = × = = = × × × (iii) The magnetic field B between the capacitor plates will be perpendicular to the electric field E and will be at right angles to it. B E E We can use Ampere’s law to determine the value of B. Since the net current (I net ) passing through the enclosed surface is zero, we have: E o o d B dl dt φ ε μ ⋅ = ∫ We can choose a circle of radius r at the center of the plate to represent the path of a magnetic field line. If E is the electric field between the two plates, the electric flux crossing the area bounded by this circle is E (...
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Lesson_4.5_Homework_and_Solutions - Lesson 4.5 Homework...

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