 Solution

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

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

For

t \to + \infty

with

qE > 0

z (t) = c t - \frac{E _{0}}{qE} + O (t^{- 1}),

so the longitudinal velocity tends to the speed of light. More generally, its limiting sign is the sign of

qE

For small

t

z (t) = \frac{qE c ^{2}}{2 E _{0}} t^{2} + O (t^{4}) .

In the nonrelativistic limit

E_{0} ≃ m c^{2}

, this becomes

z (t) ≃ \frac{qE}{2 m} t^{2} = \frac{1}{2} a t^{2}, a = \frac{qE}{m} .

The transverse motion is then uniform:

x (t) = \frac{p _{0}}{m} t .

Eliminating

t

gives the nonrelativistic parabolic trajectory

z (x) = \frac{qE m}{2 p _{0}^{2}} x^{2} .

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