Question

An electron is shot directly toward the center of a large metal plate that has surface charge density $\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathit{C}\mathbf{/}{\mathit{m}}^{\mathbf{2}}$. If the initial kinetic energy of the electron isand if the electron is$\mathbf{1}\mathbf{.}\mathbf{60}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{17}}\mathit{J}$ to stop (due to electrostatic repulsion from the plate) just as it reaches the plate, how far from the plate must the launch point be?

Short answer

The launch point must be $4.4\times {10}^{-4}$ far from the plate.

Step-by-step solution

The given data 

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

Understanding the concept of the electric field and Newton’s law

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,$E=\frac{\sigma }{{\epsilon }_0}$ (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)

Step 3: Calculation of the distance of the launch point from the plate

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:

$F=\frac{-e\sigma }{{\epsilon }_0}$

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

$a=\frac{F}{m}$……………(5)

$=\frac{-e\sigma }{m{\epsilon }_0}$

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

$x=-\frac{v_0^2}{2a}$ ………………….(5)

$=\frac{m{\epsilon }_0{v}_0^2}{2e\sigma }$

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:

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

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