 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ia/paper-3/11c/solution

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

\nabla \times (\nabla \times E) = \nabla (\nabla \cdot E) - \nabla^{2} E

\frac{\partial ^{2} E}{\partial t ^{2}} - \nabla^{2} E = - \nabla ρ - \frac{\partial J}{\partial t} .

Applying the same identity to

B

, using

\nabla \cdot B = 0

, gives

\frac{\partial ^{2} B}{\partial t ^{2}} - \nabla^{2} B = \nabla \times J .

r

gives

E (r) = \frac{r}{r ^{2}} \int_{0}^{r} ρ (s) s^{2} d s .

For the unit-density ball of radius

a

E = ⎩ ⎨ ⎧ \frac{r}{3} r, \frac{a ^{3}}{3 r ^{2}} r, r < a, r \geq a, ϕ = ⎩ ⎨ ⎧ \frac{3 a ^{2} - r ^{2}}{6}, \frac{a ^{3}}{3 r}, r < a, r \geq a .

The potential is continuous at

r = a

and tends to zero at infinity.

Let

p = (1, 0, 0)

and write

1_{S_{n}}

for the indicator of

S_{n}

. The layered density decomposes as

ρ = 1_{S_{0}} + \sum_{n \geq 1} 2^{n - 1} 1_{S_{n}} .

The point

p

lies on every sphere

S_{n}

, whose radius is

R_{n} = 4^{- n}

and whose outward radial direction there is

x

. A uniform ball of density

c

contributes

c R_{n} /3

to the field and

c R_{n}^{2} /3

to the potential at its surface. Therefore

E (p) = \frac{1}{3} (1 + \sum_{n \geq 1} 2^{n - 1} 4^{- n}) x = \frac{1}{2} x,

and

ϕ (p) = \frac{1}{3} (1 + \sum_{n \geq 1} 2^{n - 1} 1 6^{- n}) = \frac{5}{14} .

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