91Ó°ÊÓ

1. Surface charge density,$\mathrm{σ}=-2.00×{10}^{-8}C/m^2$
2. The initial kinetic energy of the electron,${\mathrm{KE}}_{\mathrm{i}}=1.60×{10}^{-17}\mathrm{J}$
3. The electron stops just reaching the plate.

Using the concept of the electric field of Gauss's flux theorem, we can get the expression of the electrostatic force. Now, the value of the force can be used to calculate the distance of the launch point from the plate.

Formulae:

The electric field at a point due to Gauss’s flux theorem,$E=\frac{\mathrm{σ}}{{\mathrm{Î\mu }}_0}$ (1)

The force due to Newton’s second law,$F=ma$ (2)

The electrostatic force on a charged body,F=qE (3)

The third equation of the kinematic motion,$v^2-v_0^2=2ax$ (4)

The charge on the metal plate, which is negative, exerts a force of repulsion on the electron and stops it.

The force on the electron is given by substituting equation (1) in equation (2) as given:

$F=\frac{-e\mathrm{σ}}{{\mathrm{Î\mu }}_0}$

The acceleration expression is given using the above equation in equation (iii) as follows:

$a=\frac{F}{m}$…â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦(5)

$=\frac{-e\mathrm{σ}}{m{\mathrm{Î\mu }}_0}$

Since, the electron rests to stop; the final velocity of the electron is given as:

Now, the expression of the distance using equation (5) in equation (4) is as follows

$x=-\frac{v_0^2}{2a}$ …â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦.(5)

$=\frac{m{\mathrm{Î\mu }}_0{v}_0^2}{2e\mathrm{σ}}$

Now $\frac{1}{2}m{v}_0^2$is the value of the initial kinetic energy$K_0$ , so the distance equation (5) is given as:

$$\begin{gathered} x=\frac{{\mathrm{Î\mu }}_0{K}_0}{e\mathrm{σ}} \\ =\frac{(8.85×{10}^{-12}{C}^2/N.m^2)(1.60×{10}^{-17}J)}{(1.60×{10}^{-19}C)(-2.00×{10}^{-6}C/m^2)} \\ =4.4×1-{10}^{-4}m \end{gathered}$$

Hence, the value of the distance of the point from the plate is $4.4×{10}^{-4}$ .