An electron moving with a speed of `5xx10^(6)m//s` is shot parallel to the electric field of strength `1xx10^(3)` N/C arranged so as to retard its motion. How far will the electron travel in the field before coming (momentarily) to rest? (`m_(e)=9.1xx10^(-31)kg`)

Options

A. 7m
B. 70m
C. 7cm
D. 0.7cm

Explanation

To solve the problem step by step, we will use the concepts of electric force, acceleration, and kinematics.

### Step 1: Identify the given values
- Speed of the electron, \( u = 5 \times 10^6 \, \text{m/s} \)
- Electric field strength, \( E = 1 \times 10^3 \, \text{N/C} \)
- Mass of the electron, \( m_e = 9.1 \times 10^{-31} \, \text{kg} \)
- Charge of the electron, \( q = -1.6 \times 10^{-19} \, \text{C} \)

### Step 2: Calculate the force acting on the electron
The force \( F \) acting on the electron due to the electric field is given by:
\[ F = qE \]
Substituting the values:
\[ F = (-1.6 \times 10^{-19} \, \text{C})(1 \times 10^3 \, \text{N/C}) = -1.6 \times 10^{-16} \, \text{N} \]
(The negative sign indicates that the force is acting opposite to the direction of motion.)

### Step 3: Calculate the acceleration of the electron
Using Newton's second law, \( F = ma \), we can find the acceleration \( a \):
\[ a = \frac{F}{m} = \frac{-1.6 \times 10^{-16} \, \text{N}}{9.1 \times 10^{-31} \, \text{kg}} \approx -1.76 \times 10^{14} \, \text{m/s}^2 \]
(The acceleration is negative, indicating that it is in the opposite direction to the motion of the electron.)

### Step 4: Use kinematic equation to find the distance traveled before coming to rest
We will use the kinematic equation:
\[ v^2 = u^2 + 2as \]
Where:
- \( v = 0 \, \text{m/s} \) (final velocity when the electron comes to rest)
- \( u = 5 \times 10^6 \, \text{m/s} \) (initial velocity)
- \( a = -1.76 \times 10^{14} \, \text{m/s}^2 \) (acceleration)
- \( s \) is the distance traveled before coming to rest.

Substituting the values:
\[ 0 = (5 \times 10^6)^2 + 2(-1.76 \times 10^{14})s \]
\[ 0 = 25 \times 10^{12} - 3.52 \times 10^{14}s \]
Rearranging gives:
\[ 3.52 \times 10^{14}s = 25 \times 10^{12} \]
\[ s = \frac{25 \times 10^{12}}{3.52 \times 10^{14}} \approx 0.07109 \, \text{m} \]

### Step 5: Convert to centimeters
To convert meters to centimeters:
\[ s \approx 0.07109 \, \text{m} \times 100 \approx 7.109 \, \text{cm} \]

### Final Answer
The distance the electron will travel before coming to rest is approximately \( 7.1 \, \text{cm} \).

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