 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/ib/paper-2/6b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

Take the electric dipole moment along the polar axis,

p = p \hat{z}

E_{r} = - \partial_{r} ϕ = \frac{2 p c o s θ}{4 π ϵ _{0} r ^{3}}, E_{θ} = - \frac{1}{r} \partial_{θ} ϕ = \frac{p s i n θ}{4 π ϵ _{0} r ^{3}}, E_{φ} = 0.

Since

p = p (cos θ \hat{e}_{r} - sin θ \hat{e}_{θ})

and

p \cdot \hat{e}_{r} = p cos θ

, the radial and angular components combine as

E (r) = \frac{3 ( p \cdot e ^ _{r} ) e ^ _{r} - p}{4 π ϵ _{0} r ^{3}}, r > 0.

The origin is excluded because the point-dipole potential is singular there.

 New to topics? Read the docs here!