Chapter 5 - Conductors, Dielectrics and Capacitance

Chapter 5 - Conductors, Dielectrics and Capacitance -...

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1 Chapter 5 Conductors, Dielectrics And Capacitance

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2 Current and Current Density Electric Current charges in motion rate of movement of charge passing a given a reference point or crossing a given reference plane unit for current is Ampere (A). 1 A = 1 C/s Current is a scalar quantity
3 Using vector notation: dI = J d S The total current passing through an area may be determined by integrating: Current density A vector representing the limit of the ratio of the current passing through a small area to the area as the area shrinks to zero Denoted by the symbol J A I lim J 0 A Δ Δ = Δ = S d I S J

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4 A coaxial cable is carrying 20 A of current. The radius of the inner conductor is 20mm while the outer conductor has an inner radius of 30mm and outer radius of 35mm. Calculate the current density through both conductors. Assume that the current is flowing uniformly through the conductor. 2 2 inner inner m kA 916 . 15 ) 02 . 0 ( 20 A I J = π = = 2 2 2 outer outer m kA 588 . 19 ) 03 . 0 035 . 0 ( 20 A I J = - π = =
5 Consider an element of charge bounded by a differential volume Δ S Δ L. During Δ t, the volume moved along the x-axis by Δ x

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6 Δ Q = ρ v Δ S Δ x has moved through a reference plane perpendicular to the direction of motion. The resultant current is t x S t Q I v Δ Δ Δ ρ = Δ Δ = Δ
7 In terms of current density: x v 0 S v S I lim J ρ = Δ Δ = Δ In general: J = ρ v v Current resulted by motion of charges is called convection current. x v v Sv I t x S t Q I Δ ρ = Δ Δ Δ Δ ρ = Δ Δ = Δ

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8 Continuity of Current Principle of conservation of charge Charges can be neither created nor destroyed, although equal amounts of positive and negative charges may be simultaneously created, by separation, or destroyed, by recombination. Consider a closed surface. The total current flowing out of the surface is = S d I S J
9 The current out of the surface must be equal to the rate at which the amount of charges inside the surface is decreasing: Applying the divergence theorem to expression on the left: dt dQ d I in S - = = S J = vol S dv ) ( d J S J

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10 ρ - = ρ - = - = vol v vol v in vol dv t dv dt d Q dt d dv ) ( J Therefore, dv t dv ) ( v ρ - = J t v ρ - = J The continuity equations: Integral form Point form dt dQ d I i S - = = S J t v ρ - = J
11 Example: Given the current density J = a r A/m 2 . a. Calculate the total current when t = 1 sec at r = 5m and r = 6m. b. Determine the volume charge density and velocity. t e r 1 - Solution: a. J (t = 1, r = 5) = e -1 /5 = 0.07358 a r A/m 2 I(r=5) = JA = 0.07358 x 4 π 5 2 = 23.115 A J (t = 1, r = 6) = e -1 /6 = 0.06131 a r A/m 2 I(r=6) = JA = 0.06131 x 4 π 6 2 = 27.737 A

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12 b.
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• Winter '10
• JoelJosephMarciano
• Electric charge, Coaxial cable, charge density, Boundary conditions, Dielectrics, volume charge density

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