A Point Charge in an Electric Field — Question 9

Question 9

Equal initial kinetic energies remove mass from a transverse-deflection formula

Two positive particles enter the same uniform transverse field E and field length L with equal initial kinetic energy K and zero transverse velocity. Their parameters are $(q_1,m_1)=(e,m)$ and $(q_2,m_2)=(2e,4m)$. Ignore edge effects and other forces.

1. Derive the exit displacement in terms of q,E,L,K.

2. Compare their exit displacements, transit times and transverse accelerations. Explain how the ratios combine.

Question 9 - Solution

Equal initial kinetic energies remove mass from a transverse-deflection formula

(a) Method and calculation

$$v_x=\sqrt{2K/m},\quad t=L/v_x,\quad y=\tfrac12(qE/m)t^2.$$$$y=\frac{qEL^2}{4K}.$$

(b) Interpretation and further calculation

$$y_2/y_1=2,\quad t_2/t_1=\sqrt{m_2/m_1}=2,\quad a_{y,2}/a_{y,1}=\frac{2e/(4m)}{e/m}=1/2.$$$$y_2/y_1=(a_{y,2}/a_{y,1})(t_2/t_1)^2=2.$$The heavier particle accelerates less but spends longer in the region.

Physical check

Mass cancellation requires equal kinetic energies, not equal entrance speeds.