Poynting Theorem Questions and Answers

Q: Which of the following is NOT a term in Poynting's theorem?

Correct answer: the total charge enclosed by the volumePoynting’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.

Q: A solar panel of 2 m² in full sunlight of 1 kW/m² and 20 per cent efficiency delivers

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.

41.

Consider the following statements about energy in a circuit.

1. Energy flows in the space around the conductors 2. The Poynting vector inside a wire accounts for its ohmic loss 3. The field and circuit pictures give different totals

A.1 and 2 only are correct
B.1, 2 and 3 are all correct statements about energy in a circuit
C.2 and 3 only are correct
D.1 and 3 only are correct

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.

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42.

Which of the following is NOT a term in Poynting's theorem?

A.the rate of change of the stored field energy
B.the total charge enclosed by the volume
C.the power crossing the bounding surface
D.the ohmic power dissipated within the volume

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.

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43.

Which statement about the far field of an antenna is NOT correct?

A.the field amplitude falls as the inverse first power of distance
B.the wave is very nearly a uniform plane wave over a small region
C.the Poynting vector there is largely reactive
D.the power density falls as the inverse square of distance

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.

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44.

An engineer computes a Poynting vector of \( 10^{6} \) W/m² in free space and reports it without comment. The proper response is

A.to accept it, such densities being ordinary in free space
B.to double it, since only half the power was counted
C.to convert it to watts by multiplying by the wavelength squared
D.to check the field, since that density implies about 19 kV/m

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.

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45.

Comparing the Poynting vector with the circuit expression for power, the two agree because

A.both are consequences of Maxwell's equations
B.the circuit expression is an approximation to the field one
C.the field expression neglects the reactive component
D.the circuit expression applies only at low frequency

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.

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46.

The momentum carried by an electromagnetic wave is related to its energy flow through

A.the intrinsic impedance of the medium
B.division by the speed of light
C.multiplication by the speed of light
D.division by the square of the speed of light

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.

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47.

The radiation pressure on a perfectly absorbing surface exposed to a power density \( S \) is

A.\( 2S/c \)
B.\( S/c^2 \)
C.\( S/c \)
D.\( Sc \)

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.

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48.

A solar panel of 2 m² in full sunlight of 1 kW/m² and 20 per cent efficiency delivers

A.2 kW
B.200 W
C.40 W
D.400 W

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.

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49.

A laser delivers 5 mW into a spot of 1 mm². The power density there is

A.5 kW/m²
B.5 W/m²
C.5 MW/m²
D.5 mW/m²

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.

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50.

The Poynting vector in a region containing only a static electric field is

A.equal to the square of the electric field divided by the impedance
B.zero, there being no magnetic field to cross it with
C.equal to the electric energy density divided by the speed of light
D.directed along the electric field lines

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.

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