 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2012/paper-2/15/c/solution

The condition at infinity gives

C_{out} = 0

. Matching the electric potential at

r = a

gives

\frac{k}{2 a} + C_{in} = - \frac{k}{a}, C_{in} = - \frac{3 k}{2 a} .

Consequently

V (r) = ⎩ ⎨ ⎧ \frac{k}{2 a ^{3}} (r^{2} - 3 a^{2}), - \frac{k}{r}, 0 \leq r < a, r \geq a .

This function is finite and differentiable at the origin. At the joining radius, both

V

and

V^{'}

match, giving

V (a) = - k / a

and

V^{'} (a) = k / a^{2}

. Thus the electric field is continuous there as well as being conservative on each side. This is the potential of a uniform spherical source, with source coefficient

3 k / a^{3}

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