Poynting Theorem Questions and Answers

Answer: C

The inward flux of the vector over the closed surface equals the dissipation within, and working it out for a cylindrical resistor returns \( VI \) as it must. It is an instructive exercise because the answer is known independently.

Answer: D

The Poynting vector is largest in the space between and around the conductors, and the conductors serve to guide it rather than to carry it. The circuit and field pictures agree on the total power but differ entirely on where it goes.

Answer: A

Correct answer: carries energy out of the capacitor's field into the resistor

The energy was stored in the electric field between the plates, and during discharge it flows out through the space around the circuit and into the resistor, where the Poynting vector points inward through the resistor’s surface. This is the picture the theorem insists on: energy travels in the field surrounding the conductors, not along inside the wires, and the wires only guide it.

Answer: B

It is a fictitious resistance in the sense that it dissipates nothing as heat, yet it absorbs power from the source exactly as a real resistance would. A short dipole has a small radiation resistance, which is why matching to it is difficult.

Answer: C

The power spreads over a spherical surface growing as the square of the radius, so the density falls accordingly and the field amplitude as the inverse first power. This is the free space path loss that every radio link budget begins with.

Answer: D

Close to the antenna the reactive component dominates and energy surges in and out each cycle; beyond a few wavelengths only the real radiating component survives. The boundary between the two regions is roughly where the measurement of a pattern becomes meaningful.

Answer: A

The power spreads uniformly over the whole sphere of area \( 4\pi r^2 \), which is the definition of an isotropic radiator. Real antennas concentrate power in some directions, and the ratio to the isotropic density is the gain.

Answer: B

Correct answer: \( S = E^2/377 \) for peak or rms consistently used

In free space \( E/H = \eta_0 = 377\ \Omega \), so \[ S = EH = \frac{E^2}{\eta_0} = \frac{E^2}{377}. \] The intrinsic impedance divides rather than multiplies, exactly as power is \( V^2/R \) in a circuit. The proviso about peak and rms matters: using rms fields gives the average power directly, while peak fields require a further factor of one half.

Answer: C

Correct answer: \( \tfrac{1}{2}\epsilon E^2 + \tfrac{1}{2}\mu H^2 \)

The electric and magnetic energy densities add, each with its factor of one half, in the same way that \( \tfrac{1}{2}CV^2 \) and \( \tfrac{1}{2}LI^2 \) add in a circuit. In a plane wave the two terms are equal at every instant, so the energy is shared evenly between the fields. The difference of the two terms and the cross product are different quantities entirely - the latter is the Poynting vector itself.

Answer: D

Correct answer: 0.265 W/m²

\[ S = \frac{E^2}{377} = \frac{10^2}{377} = 0.265\ \mathrm{W/m^2}. \] Because the field is given as rms this is directly the average power density, with no further factor of one half. Multiplying by 377 instead of dividing gives the 3770 offered, which is wrong by a factor of \( 377^2 \).