 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ib/paper-1/15d/e/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

r = 0.10 m

⟨ ∣ 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} .

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