 Solution

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

u_{E} = ϵ_{0} E^{2} /2

ℓ_{z}

of the coaxial capacitor, integrate over annular shells:

U = \int_{r_{1}}^{r_{2}} \frac{ϵ _{0}}{2} \frac{V ^{2}}{η ^{2} r ^{2}} (2 π r ℓ_{z}) d r = \frac{π ϵ _{0} ℓ _{z} V ^{2}}{η ^{2}} ln \frac{r _{2}}{r _{1}} .

Since

C = C^{'} ℓ_{z} = 2 π ϵ_{0} ℓ_{z} / η

, this becomes

U = \frac{1}{2} C V^{2} .

The integral is over the inter-cylinder electric field, and gives the stored energy, independently of which electrode is assigned zero electric potential.

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