Past exam of the mathematics course of the University of Cambridge 2002 ib Paper 1 11A Solution Created 2026-09-24 Updated 2026-10-14
Put and separate . Multiplying Laplace's equation by givesFor the regular polynomial angular modes, and is a Legendre polynomial. The radial equation has branches and . Regularity at the origin keeps only the first inside, and decay at infinity keeps only the second outside. Continuity at identifies their coefficients, so a mode hasIts outward-minus-inward radial derivative is . The prescribed angular dependence decomposes as , since and . Hence and , with no other required modes. The spherical harmonic matching with a derivative jump givesBoth branches are harmonic in their respective regions. They agree at the sphere, and their derivative jump is . The interior is regular at zero and the exterior decays at infinity. An additional homogeneous matching mode would satisfy , so its coefficient vanishes.
Resolve the angular surface charge density into Legendre polynomials:For a single degree , write the common surface potential as . Regularity and decay require radial factors inside and outside. Their exterior-minus-interior radial derivative at is . Thus spherical harmonic matching with a derivative jump gives .
Only degrees zero and two occur, yieldingBoth expressions have surface value , and their derivative difference is , checking both interface conditions. The interior quadratic is regular at the origin, and the exterior potential tends to zero.
The leading far-field term is the monopole term. Indeed the total charge is , and the normalization of the specified jump gives . No extra permittivity factor is inserted into the supplied normalization. Regularity here means finite piecewise smooth potential; its normal derivative necessarily jumps at the charged surface.