An electron is projected between two plates at separation $d=3 \mathrm{cm}$ each having length $l=11 \mathrm{cm}$ as shown in figure, at a speed of $6\times {10}^6{ \mathrm{ms}}^{-1}$ at an angle of $45°$ in a uniform field $E=2000{\mathrm{Vm}}^{-1}$ directed upwards. The electron strikes the

The electron will experience a downward acceleration a, given by

$$\begin{gathered} a=\frac{eE}{m}=\frac{1.6\times {10}^{-19}\times 2000}{9\times {10}^{-31}} \\ \Rightarrow a=\frac{eE}{m}=3.6\times {10}^{14}{ \mathrm{ms}}^{-2} \end{gathered}$$

So, we can consider the electron to be a projectile launched with some initial velocity between the plates under influence of downward acceleration $a$.

From the point of launch let us consider the rightward horizontal as the x-axis and upward vertical as y-axis, let us calculate the height to which the electron will rise and its horizontal range. Considering upward vertical motion, we have

$$\begin{gathered} 0=6\times {10}^6\sin 45°t+\frac{1}{2}\times \left(-3.5\times {10}^{14}\right)t^2 \\ \Rightarrow t=2.4\times {10}^{-8} \mathrm{s} \end{gathered}$$

Considering motion along x-axis, we get

$$\begin{gathered} x\left(\text{ range }\right)=6\times {10}^6\cos 45°\times 2.4\times {10}^{-8} \\ \Rightarrow x=0.102 \mathrm{m}=10.2 \mathrm{cm} \end{gathered}$$

The vertical displacement in half the total time i.e., at maximum height, we have

$t=\frac{T}{2}$

The vertical height is given by

$$\begin{gathered} y=\left(6\times {10}^6\sin 45°\right)\left(1.2\times {10}^{-8}\right)-\frac{1}{2}\left(3.5\times {10}^{14}\right){\left(1.2\times {10}^{-8}\right)}^2 \\ \Rightarrow y=0.0509-0.0252 \\ \Rightarrow y=0.0257=2.57 \mathrm{cm} \end{gathered}$$

So, the electron will not touch the upper plate and will strike the lower plate at the other edge.

Hence, (2) and (3) are correct.