In an old television tube, a narrow beam of electrons leaves the gun and moves along the axis of…

Step 1. Acceleration (a). $a = \dfrac{eE}{m_e} = \dfrac{1.60 \times 10^{-19}\ \text{C} \times 1.2 \times 10^{4}\ \text{N/C}}{9.11 \times 10^{-31}\ \text{kg}} = 2.11 \times 10^{15}\ \text{m/s}^2$. The field points up, so the force on the electron points down.

Step 2. Time (b). $t = \dfrac{l}{v_0} = \dfrac{0.030\ \text{m}}{2.4 \times 10^{7}\ \text{m/s}} = 1.25\ \text{ns}$.

Step 3. Displacement at the exit (c). $y_1 = -\tfrac12 at^2 = -1.65\ \text{mm}$, below the axis. It is less than $5.0\ \text{mm}$, so the electron clears the lower plate.

Step 4. Angle (d). $v_y = at = 2.63 \times 10^{6}\ \text{m/s}$, so $\tan\varphi = \dfrac{v_y}{v_0} = 0.1098$ and $\varphi = 6.26^\circ$, pointing down.

Step 5. The spot (e). After the plates the electron goes straight. In $18\ \text{cm}$ it falls a further $18\ \text{cm} \times 0.1098 = 19.76\ \text{mm}$, so the spot is at $-1.65\ \text{mm} - 19.76\ \text{mm} = -21.4\ \text{mm}$. As a check, the straight line comes from the middle of the plates: $(1.5\ \text{cm} + 18\ \text{cm}) \times 0.1098 = 21.4\ \text{mm}$ below the axis.

Step 6. Hitting a plate (f). The beam hits the lower plate when the displacement at the exit reaches $5.0\ \text{mm}$. The displacement grows in step with $E$, so $E_{\max} = 1.2 \times 10^{4}\ \text{N/C} \times \dfrac{5.0\ \text{mm}}{1.647\ \text{mm}} = 3.64 \times 10^{4}\ \text{N/C}$.