 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2009/paper-2/16/c/solution

From the preceding gradient,

\partial_{x}^{2} (r^{- 1}) = - r^{- 3} + 3 x^{2} r^{- 5}

. Summing the three Cartesian second partial derivatives gives

Δ (1/ r) = - 3 r^{- 3} + 3 (x^{2} + y^{2} + z^{2}) r^{- 5} = 0, r > 0.

Thus the reciprocal radius is harmonic away from the origin. Its negative gradient is

E = \frac{r}{r ^{3}}, \nabla \times E = 0 (r > 0) .

The curl vanishes because

E

is a gradient field and mixed second derivatives commute. The exclusion of the origin is essential: the potential and field are singular there.

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