 Solution

ID: past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2018/section-a/6/solution

Put the charged sheet at

z = 0

and the grounded plates at

z = \pm d /2

. By symmetry the electric field points away from the positive sheet with equal magnitude in the two gaps. A Gaussian pillbox crossing the sheet gives

E_{z} (0^{+}) - E_{z} (0^{-}) = \frac{σ}{ε _{0}}, ∣ E ∣ = \frac{σ}{2 ε _{0}} .

V_{s} = ∣ E ∣ \frac{d}{2} = \frac{σ d}{4 ε _{0}}, σ = 1.0 \times 1 0^{- 7} C m^{- 2}, d = 1.0 \times 1 0^{- 2} m .

ε_{0} = 8.85419 \times 1 0^{- 12} F m^{- 1}

V_{s} ≃ 28.24 V

- e

, so moving from a grounded plate to this positive potential decreases its potential energy by

e V_{s}

\frac{1}{2} m_{e} v^{2} = e V_{s} = \frac{e σ d}{4 ε _{0}}, v = \frac{e σ d}{2 ε _{0} m _{e}} ≃ 3.2 \times 1 0^{6} m s^{- 1} .

The unrounded value is

3.15 \times 1 0^{6} m s^{- 1}

, about

0.0105 c

. Either plate gives the same speed. The ideal calculation neglects the electron's disturbance of the prescribed field and image-force effects at its release point, consistently with treating it as a test particle.

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