= Solution
Let $K=eU_a$, where $U_a=100\,\mathrm V$, and let $v_z$ be the incident <electron> <speed>. Nonrelativistically, $mv_z^2=2eU_a$. 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> $a_x=eV_d/(md)$, 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 is
$$
x=\frac12a_xt_p^2+a_xt_p\frac{D-\ell/2}{v_z}
=\frac{a_x\ell D}{v_z^2}
=\frac{V_d\ell D}{2dU_a}.
$$
Thus the two screen edges require
$$
\boxed{-166.7\,\mathrm V\ \lesssim V_d\lesssim+166.7\,\mathrm V,\qquad\text{total sweep}\simeq333\,\mathrm V.}
$$
The 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 $v_x/v_z=x/D\simeq0.833$; no small-angle replacement for the trajectory was used. The <kinetic energy> remains far below $mc^2$, 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 $\pm164\,\mathrm V$; the intended thin-deflector estimate is about $\pm170\,\mathrm V$.
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