An electron is moving with speed of 1 m/s at distance of 1 m from a large sheet of charge with density σ C/m² . Find maximum value of σ such that electron hit the sheet after 1 sec. (mass of electron 9 × 10^–31 kg, permittivity of free space ε₀ = 9 × 10^–12 C² /Nm²)

The correct answer is 4.05 × 10^–22 C/m².

Step-by-Step Derivation:

1. Electric Field due to an Infinite Sheet of Charge:
The electric field $E$ at a distance from a large charged sheet with surface charge density $\sigma$ is uniform and is given by:
$E=\frac{\sigma }{2{\epsilon }_0}$

2. Electrostatic Force and Acceleration of the Electron:
An electron has a negative charge of magnitude $e=1.6\times {10}^{-19}\text{C}$. Since the sheet is positively charged, the electrostatic force pulls the electron towards the sheet:
$F=eE=\frac{e\sigma }{2{\epsilon }_0}$
Using Newton's second law, the magnitude of the acceleration $a$ of the electron (directed towards the sheet) is:
$a=\frac{F}{m}=\frac{e\sigma }{2m{\epsilon }_0}$

3. Kinematics of the Electron's Motion:
As shown in the diagram, the electron is initially at a distance $d=1\text{m}$ from the sheet. Assuming the electron is moving away from the sheet with an initial speed $u=1\text{m/s}$, the acceleration acts as a deceleration, turning it around and bringing it back to the sheet.
Let the position $x\left(t\right)$ represent the distance of the electron from the sheet at time $t$:
$x\left(t\right)=d+ut-\frac{1}{2}a{t}^2$

4. Condition to Hit the Sheet after 1 second:
For the electron to hit the sheet at $t=1\text{s}$, we set $x\left(1\right)=0$:
$0=1+\left(1\right)\left(1\right)-\frac{1}{2}a{\left(1\right)}^2$
This gives:
$2-\frac{1}{2}a=0\Rightarrow a=4{\text{m/s}}^2$
For the time of collision to be at least $1\text{s}$ (i.e., hitting after $1\text{s}$), the acceleration must satisfy $a\le 4{\text{m/s}}^2$, which corresponds to a maximum value of the surface charge density $\sigma$.

5. Calculating the Maximum Charge Density:
Using the maximum acceleration $a=4{\text{m/s}}^2$, we can solve for $\sigma$:
$\sigma =\frac{2m{\epsilon }_{0}a}{e}=\frac{2\times (9\times {10}^{-31}\text{kg})\times (9\times {10}^{-12}{\text{C}}^2/{\text{N·m}}^2)\times 4{\text{m/s}}^2}{1.6\times {10}^{-19}\text{C}}$
Simplifying the terms:
$\sigma =\frac{648\times {10}^{-43}}{1.6\times {10}^{-19}}=405\times {10}^{-24}{\text{C/m}}^2=4.05\times {10}^{-22}{\text{C/m}}^2$