In the uniform electric field of `E = 1 xx 10^(4) N C^(-1)`, an electron is accelerated from rest. The velocity of the electron when it has travelled a distance of `2 xx 10^(-2)` m is nearly
(`(e)/(m)` of electron `= 1.8 xx 10^(11) C kg^(-1)`)

Options

A. `1.6 xx 10^(6) ms^(-1)`
B. `0.85 xx 10^(6) ms^(-1)`
C. `0.425 xx 10^(6) ms^(-1)`
D. `8.5 xx 10^(6) ms^(-1)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the given values
- Electric field strength, \( E = 1 \times 10^4 \, \text{N/C} \)
- Distance travelled by the electron, \( s = 2 \times 10^{-2} \, \text{m} \)
- Charge-to-mass ratio of the electron, \( \frac{e}{m} = 1.8 \times 10^{11} \, \text{C/kg} \)

### Step 2: Calculate the acceleration of the electron
The force acting on the electron due to the electric field is given by:
\[ F = qE \]
where \( q \) is the charge of the electron. The acceleration \( a \) can be determined using Newton's second law:
\[ F = ma \implies a = \frac{F}{m} = \frac{qE}{m} \]
Substituting the charge-to-mass ratio:
\[ a = \frac{e}{m} E \]
Now substituting the known values:
\[ a = (1.8 \times 10^{11} \, \text{C/kg}) \times (1 \times 10^4 \, \text{N/C}) = 1.8 \times 10^{15} \, \text{m/s}^2 \]

### Step 3: Use kinematic equation to find the final velocity
We know the initial velocity \( u = 0 \) (since the electron starts from rest) and we need to find the final velocity \( v \) after travelling a distance \( s \). We can use the kinematic equation:
\[ v^2 = u^2 + 2as \]
Substituting the values:
\[ v^2 = 0 + 2 \cdot (1.8 \times 10^{15}) \cdot (2 \times 10^{-2}) \]
Calculating this gives:
\[ v^2 = 2 \cdot 1.8 \times 10^{15} \cdot 0.02 = 7.2 \times 10^{13} \]

### Step 4: Calculate the final velocity
Taking the square root to find \( v \):
\[ v = \sqrt{7.2 \times 10^{13}} \approx 8.49 \times 10^6 \, \text{m/s} \]
Rounding to two significant figures, we get:
\[ v \approx 8.5 \times 10^6 \, \text{m/s} \]

### Final Answer
The velocity of the electron when it has travelled a distance of \( 2 \times 10^{-2} \, \text{m} \) is nearly \( 8.5 \times 10^6 \, \text{m/s} \).

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