 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ib/paper-2/16d/c/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

With

B = 0

and

E = (0, 0, E)

p_{x} = p_{0}, p_{y} = 0, p_{z} = qEt .

E (t) = E_{0}^{2} + c^{2} q^{2} E^{2} t^{2}, E_{0} = m^{2} c^{4} + c^{2} p_{0}^{2} .

Since

v = c^{2} p / E

\overset{z}{˙} = \frac{c ^{2} qEt}{E ( t )} .

Integration from the initial position

z (0) = 0

gives

z (t) = \frac{E _{0}^{2} + c ^{2} q ^{2} E ^{2} t ^{2} - E _{0}}{qE} .

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