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Home , Displacement current, Electric flux

Physics 24100 & Optics Lecture 21 – Chapter 30 sec. 1-4

Fall 2012 Semester Matthew Jones Question

• An LC circuit has and . = 100 = 100 • If it oscillates with an amplitude of 100 mV, what is the amplitude of the current?

(a) 100 (b) 100 (c) 10 (d) 10 Question

• The energy bounces back and forth between the and the inductor. () 1 = 2 1 = 2 10 10 = = = 10 10 = 10 = 0.1 100 Maxwell’s Equations (so far)

• (Gauss’s law) ∙ = ∮ • (Gauss’s law for ) ∮ ∙ = • (Faraday’s law) ∮ ∙ ℓ = − ∙ • (Ampere’s law) ∮ ∙ ℓ = The Problem with Ampere’s Law

Current through is Current through is 0!

The current through is zero, but the electric is not zero. The changes as charge flows onto the capacitor. Maxwell’s

• We can think of the changing electric flux through as if it were a current: = = ∙ ∙ ℓ = + = + ∙ Calculating Displacement Current Parallel plate capacitor

Gauss’s law: (Lecture 5) = = Electric flux: = = Change in electric flux:

1 = = Displacement current:

= = Coaxial Capacitor

In Lecture 8 we calculated the between the inner and outer conductors using Gauss’s law:

∙ = We choose to be a cylindrical of length and radius with . < < But the left-hand-side is just so the displacement current is = = The real current flows into the ends of the Gaussian surface, the displacement current flows out the sides. Clicker Question

• Suppose charge flows between two spherical conductors. Which surface will have the largest displacement current flowing through it? (d) (a) (b)

(c) Clicker Question

• Suppose charge flows between two spherical conductors. Which surface will have the largest displacement current flowing through it? (d) (a) (b)

(c) = = Maxwell’s Equations (1864)

∙ =

∙ = ∙ ℓ = − ∙ ℓ = + Maxwell’s Equations in Free Space

In “free space” where there are no electric charges or sources of current, Maxwell’s equations are quite symmetric:

∙ =

∙ = ∙ ℓ = − ∙ ℓ = Maxwell’s Equations in Free Space

∙ ℓ = − ∙ ℓ = A changing induces an electric field. A changing electric flux induces a .

Will this process continue indefinitely? Light is an Electromagnetic Wave

• Maxwell showed that these equations contain the : 1 = where is a function of position and time. , • General solution:

, = − + ( + ) • Harmonic solutions:

, = sin ± • Wavelength, and frequency . = 2/ = /2 • Velocity, . = ⁄ = ⁄ Light is an Electromagnetic Wave

• Faraday’s Law:

∮ ∙ ℓ = − = − ∆∆

∙ ℓ = Δ − ()Δ x E x Ex Δ ≈ ∆∆ z1 z2 z By y Δ Light is an Electromagnetic Wave

• Ampere’s law:

∮ ∙ ℓ = = ∆∆

∙ ℓ = Δ − ()Δ x Ex ∆ ≈ − ∆∆ z z 1 2 z B B y y y ∆ Putting these together…

= − − = Differentiate the first with respect to : = − Differentiate the second with respect to : − =

= Velocity of Electromagnetic Waves

1 = = Speed of wave propagation is 1 = 1 = 4 × 10 / 8.854 × 10 / ∙ = . × / (speed of light) The electromagnetic spectrum

• In 1850, the only known forms electromagnetic waves were ultraviolet, visible and infrared. • The human eye is only sensitive to a very narrow range of wavelengths: Discovery of Radio Waves The Electromagnetic Spectrum Electromagnetic Waves

= • A solution is , = sin − where = = 2/ • What is the magnetic field?

= − = − cos − , = sin − Electromagnetic Waves

x

z y

• , and are mutually perpendicular. • In general, the direction is ̂ = × Energy in Electromagnetic Waves

Energy stored in electric and magnetic fields (lecture 17):

1 1 = = 2 2 For an electromagnetic wave, = ⁄ =

1 1 = = = 2 2 The total energy density is = + = Intensity of Electromagnetic Waves

• Intensity is defined as the average power transmitted per unit area. Intensity = Energy density wave velocity ×

= = = 377 Ω ≡ (Impedance of free space)

• We can construct a vector from the intensity and the direction : ̂ = × × = = = • This represents the flow of power in the direction ̂ • Average electric field: = / 2 = 2 • Units: Watts/m2 Question

• The magnetic field component of a radio wave is expressed:

= sin − • In which directions do the Poynting vector and the -field point? (a) +z,+x (b) +z,-x (c) -z,+x (d) -z,-x (e) +y,+x