 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ib/paper-2/5d/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

Let

r

be distance from the common centre. By Gauss's law, the electric field is radial and equals

E (r) = ⎩ ⎨ ⎧ 0, \frac{Q _{1}}{4 π ϵ _{0} r ^{2}} r, \frac{Q _{1} + Q _{2}}{4 π ϵ _{0} r ^{2}} r, 0 \leq r < R, R < r < 2 R, r > 2 R .

Taking the electric potential to vanish at infinity and requiring continuity across each shell gives

Φ (r) = \frac{1}{4 π ϵ _{0}} ⎩ ⎨ ⎧ \frac{Q _{1}}{R} + \frac{Q _{2}}{2 R}, \frac{Q _{1}}{r} + \frac{Q _{2}}{2 R}, \frac{Q _{1} + Q _{2}}{r}, 0 \leq r \leq R, R \leq r \leq 2 R, r \geq 2 R .

U = \frac{ϵ _{0}}{2} \int_{R^{3}} ∣ E ∣^{2} d V .

Only the two nonzero-field regions contribute, so

U = \frac{Q _{1}^{2}}{8 π ϵ _{0}} \int_{R}^{2 R} \frac{d r}{r ^{2}} + \frac{( Q _{1} + Q _{2} ) ^{2}}{8 π ϵ _{0}} \int_{2 R}^{\infty} \frac{d r}{r ^{2}} .

Therefore

U = \frac{Q _{1}^{2} + ( Q _{1} + Q _{2} ) ^{2}}{16 π ϵ _{0} R} = \frac{2 Q _{1}^{2} + 2 Q _{1} Q _{2} + Q _{2}^{2}}{16 π ϵ _{0} R} .

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