 Solution

ID: past-exam-of-the-electromagnetism-course-of-the-university-of-oxford/2023/1/c/ii/solution

Y_{00} = C

4 π ∣ C ∣^{2} = 1

, independently of its phase convention. Orthogonality to

Y_{00}

gives

\int_{S^{2}} \overline{Y_{ℓ m} (Ω)} d Ω = 0 (ℓ > 0), \int_{S^{2}} \overline{Y_{00} (Ω)} Y_{00} (Ω_{0}) d Ω = 1.

Only the

ℓ = m = 0

term contributes to the surface integral of the surface charge density. Multiplying by the surface-area factor

a^{2}

gives

Q_{a} = - q a^{2} \frac{a ^{- 1}}{1 - a / b} (\frac{1}{r _{0}} - \frac{1}{b}) .

Q_{a} = - q \frac{a ( b - r _{0} )}{r _{0} ( b - a )} .

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