An electron is fired directly towards the center of a large metal plate that has excess negative charge with surface charge density `=2.0xx10^(-6) C//m^(2)`. If the initial kinetic energy of electron is 100 eV and if it is to stop due to repulsion just as it reaches the plate, how far from the plate must it be fired ?

Options

A. 0.44 mm
B. 0.20 mm
C. 1 mm
D. 0.30 mm

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Problem
We have an electron that is fired towards a negatively charged metal plate. The initial kinetic energy of the electron is given as 100 eV, and the surface charge density of the plate is \( \sigma = 2.0 \times 10^{-6} \, \text{C/m}^2 \). We need to find the distance from which the electron must be fired so that it stops just as it reaches the plate.

### Step 2: Convert Kinetic Energy to Joules
The initial kinetic energy (KE) of the electron is given in electron volts (eV). We need to convert this to joules (J) using the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \).

\[ KE = 100 \, \text{eV} = 100 \times 1.6 \times 10^{-19} \, \text{J} = 1.6 \times 10^{-17} \, \text{J} \]

### Step 3: Calculate the Electric Field (E) due to the Plate
The electric field \( E \) created by an infinite plane sheet with surface charge density \( \sigma \) is given by:

\[ E = \frac{\sigma}{\epsilon_0} \]

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:

\[ E = \frac{2.0 \times 10^{-6}}{8.85 \times 10^{-12}} \approx 2.26 \times 10^{5} \, \text{N/C} \]

### Step 4: Relate Work Done to Change in Kinetic Energy
According to the work-energy theorem, the work done by the electric field on the electron as it moves towards the plate is equal to the change in kinetic energy. Since the electron comes to a stop, the work done is equal to the initial kinetic energy:

\[ W = KE = F \cdot s \]

where \( F \) is the force on the electron and \( s \) is the distance from which it is fired.

### Step 5: Calculate the Force (F) on the Electron
The force \( F \) on the electron due to the electric field is given by:

\[ F = qE \]

where \( q \) is the charge of the electron, \( q = -1.6 \times 10^{-19} \, \text{C} \). Thus,

\[ F = -1.6 \times 10^{-19} \times 2.26 \times 10^{5} \approx -3.62 \times 10^{-14} \, \text{N} \]

### Step 6: Substitute into the Work Equation
Now substituting \( F \) into the work equation:

\[ 1.6 \times 10^{-17} = 3.62 \times 10^{-14} \cdot s \]

### Step 7: Solve for Distance (s)
Rearranging the equation to find \( s \):

\[ s = \frac{1.6 \times 10^{-17}}{3.62 \times 10^{-14}} \approx 4.43 \times 10^{-4} \, \text{m} \]

### Step 8: Convert to Millimeters
To convert meters to millimeters, we multiply by 1000:

\[ s \approx 4.43 \times 10^{-4} \, \text{m} \times 1000 = 0.443 \, \text{mm} \]

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

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