 Solution

ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2013-resit/9/c/solution

For

θ = 3 0^{\circ}

, the result in b gives

1/3 = 2 E_{0} / (E + E_{0})

, hence

E = 5 E_{0}

. Acceleration from rest supplies only the change in kinetic energy:

e ∣Δ V ∣ = E - E_{0} = 4 E_{0}, ∣Δ V ∣ = \frac{4 m _{e} c ^{2}}{e} ≃ 2.044 MV .

The accelerating electric potential difference is a magnitude; its sign depends on which electrode is taken as the reference, since the electron has negative electric charge.

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