An electron entering field normally with a velocity `4 xx 10^7 ms^(-1)` travels a distance of 0.10 m in an electric field of intensity `3200 Vm^(-1)`. What is the deviation from its path?

Options

A. `1.76 mm`
B. `17.6 mm`
C. `176 mm`
D. `0.176 mm`

Answer

To solve the problem of finding the deviation of an electron entering an electric field, we can follow these steps:

### Step 1: Identify the Given Values
- Velocity of the electron, \( v = 4 \times 10^7 \, \text{m/s} \)
- Distance traveled in the electric field, \( x = 0.10 \, \text{m} \)
- Electric field intensity, \( E = 3200 \, \text{V/m} \)
- Charge of the electron, \( e = 1.6 \times 10^{-19} \, \text{C} \)
- Mass of the electron, \( m = 9.1 \times 10^{-31} \, \text{kg} \)

### Step 2: Calculate the Force on the Electron
The force \( F \) acting on the electron in the electric field is given by:
\[ F = Q \cdot E = -e \cdot E \]
Since the electron has a negative charge, the force will act in the opposite direction to the electric field. The magnitude of the force is:
\[ F = 1.6 \times 10^{-19} \, \text{C} \times 3200 \, \text{V/m} = 5.12 \times 10^{-16} \, \text{N} \]

### Step 3: Calculate the Acceleration of the Electron
Using Newton's second law, the acceleration \( a \) of the electron can be calculated as:
\[ a = \frac{F}{m} = \frac{5.12 \times 10^{-16} \, \text{N}}{9.1 \times 10^{-31} \, \text{kg}} \approx 5.62 \times 10^{14} \, \text{m/s}^2 \]

### Step 4: Calculate the Time of Travel in the Electric Field
The time \( t \) taken to travel the distance \( x \) can be calculated using the formula:
\[ t = \frac{x}{v} = \frac{0.10 \, \text{m}}{4 \times 10^7 \, \text{m/s}} = 2.5 \times 10^{-9} \, \text{s} \]

### Step 5: Calculate the Vertical Displacement (Deviation)
The vertical displacement \( y \) due to the acceleration during the time \( t \) can be calculated using the formula for uniformly accelerated motion:
\[ y = \frac{1}{2} a t^2 \]
Substituting the values:
\[ y = \frac{1}{2} \cdot 5.62 \times 10^{14} \, \text{m/s}^2 \cdot (2.5 \times 10^{-9} \, \text{s})^2 \]
Calculating \( t^2 \):
\[ t^2 = (2.5 \times 10^{-9})^2 = 6.25 \times 10^{-18} \, \text{s}^2 \]
Now substituting back:
\[ y = \frac{1}{2} \cdot 5.62 \times 10^{14} \cdot 6.25 \times 10^{-18} \approx 1.76 \times 10^{-3} \, \text{m} = 1.76 \, \text{mm} \]

### Final Answer
The deviation from its path is approximately \( 1.76 \, \text{mm} \).

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