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Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 1 / 15D / e / Solution

For isotropic radiation, the average power is spread over a sphere. At

r = 0.10 m

, the intensity of an isotropic radiator is

⟨ ∣ S ∣ ⟩ = \frac{P}{4 π r ^{2}} = \frac{1}{4 π ( 0.10 ) ^{2}} ≃ 7.96 W m^{- 2} .

Using the result of part (d),

∣ E_{0} ∣ = \frac{2 ⟨ ∣ S ∣ ⟩}{ϵ _{0} c} ≃ 77 V m^{- 1} .

Solved by gpt-5.6-sol high.

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