Problem#1
The electric flux through a certain area of a dielectric is (8.76 x 103 Vm/s4)t4. The displacement current through that area is 12.9 pA at time t = 26.1 ms. Calculate the dielectric constant for the dielectric.Answer:
The electric flux through a certain area of a dielectric is ΦE = 8.76 x 103 Vm/s4)t4, then
dφE/dt = 4(8.76 x 103 Vm/s4)t3
Because
dΦE/dt = εiD
4(8.76 x 103 Vm/s4)t3 = ε(12.9 x 10-12 A)
At t = 26.1 ms, then
4(8.76 x 103 Vm/s4)(26.1 x 10-3 s)3 = ε(12.9 x 10-12 A)
ε = 2.07 x 10-11 F/m,
so that
ε = Kε0, with ε0 = 8.854 x 10-12 F/m
so, K = 2.34
Problem#2
A parallel-plate, air-filled capacitor is being charged as in Fig. 1. The circular plates have radius 4.00 cm, and at a particular instant the conduction current in the wires is 0.280 A. (a) What is the displacement current density jD in the air space between the plates? (b) What is the rate at which the electric field between the plates is changing? (c) What is the induced magnetic field between the plates at a distance of 2.00 cm from the axis? (d) At 1.00 cm from the axis?
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| Fig.1 |
Answer:
(a) From Eqs. iC = dq/dt = ε dφE/dt and iD = ε dφE/dt
show that iC = iD and also relate iD to the rate of change of the electric field flux between the plates. Use this to calculate dE/dt and apply the generalized form of Ampere’s law ∮B.dl = µ0(iC + iD)encl to calculate B. So,
jD = iD/A = iC/A = 0.280 A/[π x (0.0400 m)2] = 55.7 A/m2
(b) the rate at which the electric field between the plates is changing is
jD = ε0dE/dt
dE/dt = jD/ε0 = 55.7 A/m2/(8.854 x 10-12 F/m) = 6.29 x 10-2 V/m.s
(c) Apply Ampere’s law
∮B.dl = µ0(iC + iD)encl
B(2πr) = µ0(iC + iD)encl
With iC = 0 (no conduction current flows through the air space between the plates) The displacement current enclosed by the path is jDπr2.
Thus
B(2πr) = µ0 jDπr2
B = ½ µ0 jDr
So, the induced magnetic field between the plates at a distance of 2.00 cm from the axis is
B = ½ (4π x 10-7 T.m/A)( 55.7 A/m2)(0.0200 m) = 7.00 x 10-7 T
(d) the induced magnetic field between the plates at a distance of 1.00 cm from the axis is
B = ½ (4π x 10-7 T.m/A)( 55.7 A/m2)(0.0100 m) = 3.50 x 10-7 T
The definition of displacement current allows the current to be continuous at the capacitor. The magnetic field between the plates is zero on the axis ( 0) r = and increases as r increases.
Problem#3
Displacement Current in a Dielectric. Suppose that the parallel plates in Fig. 1 have an area of 3.00 cm2 and are separated by a 2.50-mm-thick sheet of dielectric that completely fills the volume between the plates. The dielectric has dielectric constant 4.70. (You can ignore fringing effects.) At a certain instant, the potential difference between the plates is 120 V and the conduction current equals 6.00 mA. At this instant, what are (a) the charge q on each plate; (b) the rate of change q of charge on the plates; (c) the displacement current in the dielectric?
Answer:
(a) the charge q on each plate is
q = CV. For a parallel-plate capacitor, C = εA/d. Then,
q = (εA/d)V = (4.70)(ε0)(3.00 x 10-4 m2)(120 V)/(2.50 x 10-2 m) = 5.99 x 10-10 C
(b) the rate of change q of charge on the plates is
dq/dt = iC = 6.00 mA
(c) the displacement current in the dielectric is
jD = εdE/dt = Kε0(iC/Kε0A) = iC/A = jC
so, iD = 6.00 mA
iD = iC, so Kirchhoff’s junction rule is satisfied where the wire connects to each capacitor plate.
Problem#4
In Fig. 1 the capacitor plates have area 5.00 cm2 and separation 2.00 mm. The plates are in vacuum. The charging current iC has a constant value of 1.80 mA. At the charge on the plates is zero. (a) Calculate the charge on the plates, the electric field between the plates, and the potential difference between the plates when t = 0.500 µs. (b) Calculate dE/dt the time rate of change of the electric field between the plates. Does dE/dt vary in time? (c) Calculate the displacement current density jD between the plates, and from this the total displacement current iD. How do and compare?
Answer:
(a) the charge on the plates, the electric field between the plates, and the potential difference between the plates when t = 0.500 µs is
q = iCt = (1.80 x 10-3 A)(0.500 x 10-6 s) = 9.00 x 10-10 C
E = σ/A = q/Aε0 = 9.00 x 10-10 C/(5.00 x 10-4 m2 x 8.854 x 10-12 F/m) = 2.03 x 105 V/m
V = Ed = (2.03 x 105 V/m)(2 x 10-3 m) = 4.06 x 102 V
(b) dE/dt the time rate of change of the electric field between the plates is
dE/dt = (dq/dt)/(ε0A) = iC/ε0A
dE/dt = (1.80 x 10-3 A)/(8.854 x 10-12 F/m x 5.00 x 10-4 m2) = 4.07 x 1011 V/m.s
Since iC is constant dE/dt does not vary in time.
(c) the displacement current density jD between the plates, and from this the total displacement current iD is
jD = ε0dE/dt = (8.854 x 10-12 F/m)(4.07 x 1011 V/m.s) = 3.60 A/m2
iD = jDA = (3.60 A/m2)(5.00 x 10-4 m2) = 1.80 x 10-3 A. (iD = iC)
iD = iC . The constant conduction current means the charge q on the plates and the electric field between them both increase linearly with time and iD is constant.
Problem#5
Displacement Current in a Wire. A long, straight, copper wire with a circular cross-sectional area of 2.1 mm2 carries a current of 16 A. The resistivity of the material is 2.0 x 10-8 Ω.m (a) What is the uniform electric field in the material ? (b) If the current is changing at the rate of 4000 A/s at what rate is the electric field in the material changing? (c) What is the displacement current density in the material in part (b)? (Hint: Since K for copper is very close to 1, use ε = ε0. (d) If the current is changing as in part (b), what is the magnitude of the magnetic field 6.0 cm from the center of the wire? Note that both the conduction current and the displacement current should be included in the calculation of B. Is the contribution from the displacement current significant?
Answer:
Ohm’s law says E = ρJ. Apply Ohm’s law to a circular path of radius r.
(a) the uniform electric field in the material is
E = ρJ = ρi/A = (2.0 x 10-8 Ω.m)(16 A)/(2.1 x 10-6 m2) = 0.15 V/m
(b) the rate is the electric field in the material changing is
dE/dt = d(ρi/A)/dt = (ρ/A)di/dt
dE/dt = (2.0 x 10-8 Ω.m)(4000 A/s)/(2.10 x 10-6 m2) = 38 V/m.s
(c) the displacement current density in the material is
jD = ε0dE/dt = (8.854 x 10-12 F/m)(38 V/m.s) = 3.4 x 10-10 A/m2
(d) iD = jDA = (3.4 x 10-10 A/m2)(2.1 x 10-6 m2) = 7.14 x 10-16 A. Then
BD = µ0iD/2πr = (4π x 10-7 T.m/A)(7.14 x 10-16 A)/(2π x 0.060 m) = 2.38 x 10-21 T,
and this is a negligible contribution.
BC = µ0iC/2πr = (4π x 10-7 T.m/A)(16 A)/(2π x 0.060 m) = 5.33 x 10-5 T,
In this situation the displacement current is much less than the conduction current.
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