From what distance should a 100 eV electron be fired towards a large metal plate having a surface charge of `-2.0 xx 10^(-6) Cm^(-2)`, so that it just fails to strike the plate ?

Options

A. 0.50 mm
B. 0.44 m
C. 0.60 mm
D. 0.77 mm

Answer

To solve the problem step by step, we will use the concepts of electric field, potential difference, and energy conservation.

### Step 1: Understand the Given Data
We have:
- Energy of the electron, \( E = 100 \, \text{eV} \)
- Surface charge density of the plate, \( \sigma = -2.0 \times 10^{-6} \, \text{C/m}^2 \)

### Step 2: Convert Energy from eV to Joules
1 eV is equal to \( 1.6 \times 10^{-19} \, \text{J} \). Therefore, we convert the energy of the electron:
\[ E = 100 \, \text{eV} = 100 \times 1.6 \times 10^{-19} \, \text{J} = 1.6 \times 10^{-17} \, \text{J} \]

### Step 3: Calculate the Electric Field (E) due to the Charged Plate
The electric field \( E \) due to a large charged plate is given by the formula:
\[ E = \frac{\sigma}{\epsilon_0} \]
where \( \epsilon_0 \) (the permittivity of free space) is approximately \( 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \).

Substituting the values:
\[ E = \frac{-2.0 \times 10^{-6}}{8.85 \times 10^{-12}} \approx -226.5 \, \text{N/C} \]

### Step 4: Relate Electric Field to Potential Difference
The potential difference \( V \) between two points in an electric field is given by:
\[ V = E \cdot d \]
where \( d \) is the distance from the plate.

### Step 5: Relate Energy to Work Done
The work done on the electron when it moves through a potential difference \( V \) is equal to its energy:
\[ W = qV \]
where \( q \) is the charge of the electron, approximately \( -1.6 \times 10^{-19} \, \text{C} \).

Setting the work done equal to the energy of the electron:
\[ 1.6 \times 10^{-17} = (-1.6 \times 10^{-19}) \cdot V \]

### Step 6: Substitute for V
From the previous equation, we can express \( V \) as:
\[ V = \frac{1.6 \times 10^{-17}}{-1.6 \times 10^{-19}} = -100 \, \text{V} \]

### Step 7: Substitute V into the Electric Field Equation
Now, substituting \( V \) back into the electric field equation:
\[ -100 = (-226.5) \cdot d \]
Solving for \( d \):
\[ d = \frac{-100}{-226.5} \approx 0.441 \, \text{m} \]

### Step 8: Convert Distance to Millimeters
To convert meters to millimeters:
\[ d \approx 0.441 \, \text{m} \times 1000 \, \text{mm/m} = 441 \, \text{mm} \]

### Step 9: Final Answer
The distance from which the electron should be fired is approximately \( 0.44 \, \text{mm} \).

### Conclusion
Thus, the answer is \( 0.44 \, \text{mm} \), which corresponds to option 2.