Past exam of the physics 2 paper of the University of Oxford / 2017 / 1 / Solution

Let $K=e{U}_a$, where $U_a=100\mathrm{V}$, and let $v_z$ be the incident electron speed. Nonrelativistically, $m{v}_z^2=2e{U}_a$. Take the plate length along the beam as $\ell =0.01\mathrm{m}$, their separation as $d=0.01\mathrm{m}$, and their centre-to-screen distance as $D=0.30\mathrm{m}$. A signed deflecting voltage $V_d$ gives the transverse acceleration $a_x=e{V}_d/(md)$, choosing its sign so that positive $V_d$ deflects the electron towards positive $x$. The transit time is $t_p=\ell /v_z$. The displacement inside the plates and in the subsequent field-free drift isThus the two screen edges requireThe transverse deflection at plate exit is $x_p=x\ell /(2D)=4.17\mathrm{mm}$ at either extreme, smaller than the half-gap $5\mathrm{mm}$, so the beam clears the plates. The outgoing slope is $v_x/v_z=x/D\simeq 0.833$; no small-angle replacement for the trajectory was used. The kinetic energy remains far below $m{c}^2$, so the nonrelativistic approximation is adequate. If “30 cm behind” is instead measured from the downstream plate edge, replace $D$ by $0.305\mathrm{m}$, giving approximately $\pm 164\mathrm{V}$; the intended thin-deflector estimate is about $\pm 170\mathrm{V}$.