Electromagnetic Waves: questions with answers

Current in a wire deposits or removes plate charge, changing the gap field. Maxwell displacement current accounts for that changing flux; it is not electron transport across the vacuum.

Displacement-current density is epsilon0 times the time derivative of E. A field can be nonzero or vary with position while its time derivative is zero.

Radio is an electromagnetic wave of fields and needs no material medium. Ordinary sound requires material particles to carry pressure disturbances.

Electromagnetic waves are transverse and need no medium, but transversality alone does not imply medium independence: mechanical transverse waves still require matter.

Propagation is along E cross B. With B to the right, upward E gives into-page propagation and downward E gives out-of-page propagation.

For propagation along +z with E along x, use B along y and the same sinusoidal phase. A cosine introduces a quarter-cycle shift and fails the travelling-wave relation.

Both travel at c. Since lambda=c/nu, the larger gamma frequency corresponds to a shorter wavelength rather than a greater vacuum speed.

The spatial phase coefficient is the wave number k. One wavelength changes phase by 2π, so lambda=2π/k.

Use u_E=epsilon0 E²/2 and u_B=B²/(2mu0). With E=cB and c²=1/(mu0 epsilon0), the contributions are equal, also after averaging.

The low-frequency visible end is red. Beyond it towards longer wavelengths lies infrared; ultraviolet is beyond the higher-frequency violet end.

The chapter explicitly treats the spectrum boundaries as approximate with overlaps. In this region atomic/electron versus nuclear origin can be needed for classification.

For the same aperture, shorter wavelength reduces diffraction. Since nu lambda=c in vacuum, increasing frequency achieves that shorter wavelength.

Nuclear de-excitation is a gamma-ray source. X-rays are commonly associated with energetic electron targets or atomic inner-shell transitions.

For a chosen spanning surface, the current in Ampère–Maxwell’s law is conduction current plus $\varepsilon_0\,d\Phi_E/dt$ through that same surface.

In the ideal capacitor model, a surface cutting the lead carries conduction current i, whereas an alternative surface through the gap carries displacement current $i_d=dQ/dt=i$.

Each surface therefore gives the same total current as the lead: ratio 1 : 1. Do not add the currents from two alternative spanning surfaces.

Conduction and displacement currents are added only when evaluated through one and the same chosen surface.

In the ideal capacitor model, the wire-cutting surface has conduction current $i_c$; the alternative gap surface has displacement current $i_d=i_c$. Both give the same source current.

The student adds values from different alternative spanning surfaces and double-counts the same current. The source current is $i_c$, not $2i_c$.