 Solution

ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2017/8/a/solution

V = ϕ (r_{2}) - ϕ (r_{1})

and

η = ln (r_{2} / r_{1}) > 0

. Cylindrical symmetry and Gauss's law imply a radial electric field proportional to

1/ r

. Integrating

E = - \nabla ϕ

fixes the sign:

ϕ (r) - ϕ (r_{1}) = \frac{V l n ( r / r _{1} )}{η}, E (r) = - \frac{V}{ηr} \hat{r} .

λ_{q}

2 π r E_{r} = λ_{q} / ϵ_{0}

, so

λ_{q} = - 2 π ϵ_{0} V / η

. Capacitance is positive regardless of the chosen terminal polarity:

C^{'} = \frac{∣ λ _{q} ∣}{∣ V ∣} = \frac{2 π ϵ _{0}}{ln ( r _{2} / r _{1} )} .

End effects are neglected for the long cylinders, as required by the cylindrical approximation.

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