91Ó°ÊÓ

When a charge (q) at a point (p) acting upon an electric field (F)

$E=\frac{F}{q}$

As the electric force is$F=m_{proton}.a$

$E=\frac{F=m_{proton}.a}{q}$

As role="math" localid="1668169647542" $m_{proton}=1.67×{10}^{-27}kg,q_{proton}=1.60×{10}^{-19}C$

At first, calculating acceleration using the final velocity expression

$v^2=v_0^2+2ad$, as v=0

$$\begin{gathered} a=\frac{v_0^2}{2d} \\ =\frac{{\left(4.5×{10}^6\right)}^2}{2×0.032} \\ =-3.164×{10}^{14}m/s^2 \end{gathered}$$

Putting the value to determine the electric field

$$\begin{gathered} E=\frac{1.67×{10}^{-27}×-3.164×{10}^{14}}{1.60×{10}^{-19}} \\ E=-3.3×{10}^{6}N/C \end{gathered}$$

Hence the magnitude and direction of the weakest electric field is$-3.3×{10}^{6}N/C$.

Time required by proton

$$\begin{gathered} v=v_0+at;v=0 \\ t=-\frac{v_0}{a} \\ =\frac{-4.5×{10}^6}{-3.164×{10}^{14}} \\ =1.422×{10}^{-8}s \end{gathered}$$

Therefore, Time taken by the proton after entering the field is $1.422×{10}^{-8}s$.

$$\begin{gathered} m_{electron}=9.202×{10}^{-31}kg \\ {q}_{electron}=-1.60×{10}^{-19}C \end{gathered}$$

Putting all values in the equation

$$\begin{gathered} E=\frac{m_{electron}.a}{q_{electron}} \\ =\frac{\left(9.202×{10}^{-31}\right)\left(-3.164×{10}^{14}\right)}{-1.60×{10}^{-19}} \\ =1.8×{10}^{3}N/C \end{gathered}$$

Hence, the minimum field required to stop electrons is$1.8×{10}^{3}N/C$ and direction is towards the direction to the speed of electron.