Past exam of the physics 2 paper of the University of Oxford / 2021 resit / 1 / c

Solution

\pm Q

over a common length

L

λ = Q / L

ℓ

give

E (r) 2 π r ℓ = \frac{λ ℓ}{ϵ _{0}}, E (r) = \frac{λ}{2 π ϵ _{0} r}, a < r < b .

V = \int_{a}^{b} E (r) d r = \frac{Q}{2 π ϵ _{0} L} lo g \frac{b}{a}, C = \frac{2 π ϵ _{0} L}{lo g ( b / a )} .

This neglects the fringing field of a capacitor at the ends, as justified when

L ≫ b

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