A particle of mass `10^(-3)`kg and charge `5muC` is thrown at a speed of `20ms^(-1)` against a uniform electric field of strength `2xx10^(5)NC^(-1)`. The distance travelled by particle before coming to rest is

Options

A. 0.1 m
B. 0.2 m
C. 0.3 m
D. 0.4 m

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the given values
- Mass of the particle, \( m = 10^{-3} \, \text{kg} \)
- Charge of the particle, \( q = 5 \, \mu\text{C} = 5 \times 10^{-6} \, \text{C} \)
- Initial speed of the particle, \( u = 20 \, \text{m/s} \)
- Electric field strength, \( E = 2 \times 10^{5} \, \text{N/C} \)

### Step 2: Calculate the force acting on the particle
The force \( F \) acting on the charged particle in the electric field is given by:
\[ F = E \cdot q \]
Substituting the values:
\[ F = (2 \times 10^{5} \, \text{N/C}) \cdot (5 \times 10^{-6} \, \text{C}) = 1 \, \text{N} \]

### Step 3: Determine the acceleration of the particle
Since the particle is thrown against the electric field, the force will act in the opposite direction to the motion. Therefore, the acceleration \( a \) can be calculated using Newton's second law:
\[ F = m \cdot a \implies a = -\frac{F}{m} \]
Substituting the values:
\[ a = -\frac{1 \, \text{N}}{10^{-3} \, \text{kg}} = -10^{3} \, \text{m/s}^2 \]

### Step 4: Use the kinematic equation to find the distance traveled
We can use the kinematic equation:
\[ v^2 = u^2 + 2as \]
Where:
- \( v \) is the final velocity (0 m/s, since the particle comes to rest),
- \( u \) is the initial velocity (20 m/s),
- \( a \) is the acceleration (-1000 m/s²),
- \( s \) is the distance traveled.

Rearranging the equation to solve for \( s \):
\[ 0 = (20)^2 + 2(-10^{3})s \]
\[ 0 = 400 - 2000s \]
\[ 2000s = 400 \]
\[ s = \frac{400}{2000} = 0.2 \, \text{m} \]

### Final Answer
The distance traveled by the particle before coming to rest is \( 0.2 \, \text{m} \).

---