Poynting Theorem Questions and Answers

Answer: A

The first two are correct and are the substance of the field picture. The two pictures agree exactly on the total power, differing only on where it is regarded as travelling, which is why circuit theory serves so well.

Answer: B

Correct answer: the total charge enclosed by the volume

Poynting’s theorem is a statement of energy conservation, and its three terms are the power crossing the bounding surface, the rate of change of stored field energy inside, and the power dissipated inside. Charge does not appear; it belongs to Gauss’s law, which is a statement about flux rather than about energy.

Answer: C

Correct answer: the Poynting vector there is largely reactive

That describes the *near* field, not the far field. Far from an antenna the vector is almost entirely real, since that is what radiation means. The other three statements are all correct: the field amplitude falls as \( 1/r \), the power density therefore as \( 1/r^2 \), and over a small region the wavefront is flat enough to be treated as a uniform plane wave.

Answer: D

A megawatt per square metre corresponds to a field of nineteen kilovolts per metre, which is well within what air will sustain but far above any ordinary radio signal. It would be reasonable inside a high power radar waveguide and absurd for a broadcast field.

Answer: A

Integrating the vector over a cross-section of a transmission line returns the product of the voltage and current exactly, with no approximation. Circuit theory is a limiting case of field theory but the power expressions coincide identically wherever both apply.

Answer: B

Dividing the Poynting vector by the speed of light gives the momentum density, which is why light exerts a pressure on a surface it strikes. That radiation pressure is small but measurable and is the basis of the solar sail.

Answer: C

The momentum arriving per unit area per unit time is the energy flux divided by the speed of light, and absorbing it delivers that momentum. A perfect reflector receives twice as much, since the momentum is reversed rather than merely stopped.

Answer: D

Correct answer: 400 W

\[ 2\ \mathrm{m^2} \times 1000\ \mathrm{W/m^2} \times 0.20 = 400\ \mathrm{W}. \] The power density is multiplied by the area to give the power intercepted, 2 kW, and then by the efficiency to give what is delivered. Applying the efficiency without the area, or the area without the efficiency, produces two of the distractors offered.

Answer: A

A square millimetre is \( 10^{-6} \) m², so the density is \( 5 \times 10^{-3}/10^{-6} = 5 \times 10^{3} \) W/m². Concentrating a modest power into a small area is what makes even a low-power laser dangerous to the eye.

Answer: B

Both fields must be present for the cross product to be non-zero, so a purely electrostatic region has no energy flow. Energy is stored there but nothing is transported, which is exactly what one would expect of a static situation.