 Solution

ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2013/11/i/solution

E = σ / ϵ_{0}

between the parallel plates. Their voltage is

V = E a = σa / ϵ_{0}

, and their charge magnitude is

Q = σ s^{2}

. Therefore

C_{0} = \frac{Q}{V} = \frac{ϵ _{0} s ^{2}}{a} .

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