A charged particle having charge q and mass m moves rectilinearly under the action of the electric field `E(x)=(4-3x)N//C` where x is the distance was initially at rest. The distance travelled by the particle till it comes instantaneously to rest again for the first time, is

Options

A. `(2)/(3)m`
B. `(4)/(3)m`
C. 2m
D. `(8)/(3)m`

Answer

To solve the problem, we need to determine the distance traveled by a charged particle until it comes to rest for the first time while moving under the influence of a varying electric field \( E(x) = 4 - 3x \) N/C.

### Step-by-Step Solution:

1. **Understanding the Force Acting on the Particle**:
The force \( F \) acting on the charged particle is given by:
\[ F = qE(x) = q(4 - 3x) \]

2. **Using Newton's Second Law**:
According to Newton's second law, the force can also be expressed as:
\[ F = ma \]
where \( a \) is the acceleration of the particle. Therefore, we can equate the two expressions for force:
\[ ma = q(4 - 3x) \]
This gives us the acceleration:
\[ a = \frac{q}{m}(4 - 3x) \]

3. **Relating Acceleration to Velocity**:
Since acceleration \( a \) can be expressed as \( a = \frac{dv}{dt} \) and using the chain rule, we can write:
\[ a = \frac{dv}{dx} \cdot v \]
Thus, we can set up the equation:
\[ v \frac{dv}{dx} = \frac{q}{m}(4 - 3x) \]

4. **Separating Variables**:
We can rearrange this to separate the variables:
\[ v dv = \frac{q}{m}(4 - 3x) dx \]

5. **Integrating Both Sides**:
We will integrate both sides. The left side integrates from \( 0 \) to \( v \) (initial velocity \( u = 0 \)):
\[ \int_0^v v \, dv = \frac{1}{2} v^2 \]
The right side integrates from \( 0 \) to \( x \):
\[ \int_0^x \frac{q}{m}(4 - 3x) \, dx = \frac{q}{m} \left( 4x - \frac{3}{2}x^2 \right) \bigg|_0^x = \frac{q}{m} \left( 4x - \frac{3}{2}x^2 \right) \]

6. **Setting the Integrals Equal**:
Setting the two integrals equal gives:
\[ \frac{1}{2} v^2 = \frac{q}{m} \left( 4x - \frac{3}{2}x^2 \right) \]

7. **Finding the Condition for Coming to Rest**:
The particle comes to rest again when \( v = 0 \). Thus, we set the left side to zero:
\[ 0 = \frac{q}{m} \left( 4x - \frac{3}{2}x^2 \right) \]
This implies:
\[ 4x - \frac{3}{2}x^2 = 0 \]
Factoring out \( x \):
\[ x(4 - \frac{3}{2}x) = 0 \]
This gives us two solutions:
\[ x = 0 \quad \text{or} \quad 4 - \frac{3}{2}x = 0 \]

8. **Solving for the Non-zero Solution**:
Solving \( 4 - \frac{3}{2}x = 0 \):
\[ \frac{3}{2}x = 4 \implies x = \frac{8}{3} \text{ meters} \]

### Final Answer:
The distance traveled by the particle until it comes instantaneously to rest again for the first time is:
\[ \boxed{\frac{8}{3} \text{ meters}} \]