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 of a charged particle moving in a variable electric field, we will follow these steps:

### Step 1: Understand the Given Information
We have a charged particle with charge \( q \) and mass \( m \) moving under the influence of an electric field given by:
\[ E(x) = 4 - 3x \, \text{N/C} \]
The particle starts from rest at \( x = 0 \) and we need to find the distance traveled by the particle until it comes to rest again.

### Step 2: Relate Force, Mass, and Acceleration
The force acting on the particle due to the electric field is given by:
\[ F = qE = q(4 - 3x) \]
According to Newton's second law, we have:
\[ F = ma \]
Thus, we can write:
\[ ma = q(4 - 3x) \]
Rearranging gives us the expression for acceleration \( a \):
\[ a = \frac{q}{m}(4 - 3x) \]

### Step 3: Express Acceleration in Terms of Velocity and Position
Acceleration can also be expressed as:
\[ a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx} \]
Substituting this into our equation gives:
\[ v \frac{dv}{dx} = \frac{q}{m}(4 - 3x) \]

### Step 4: Rearranging and Integrating
We can rearrange the equation to integrate:
\[ v \, dv = \frac{q}{m}(4 - 3x) \, dx \]
Now, we will integrate both sides. The left side integrates from \( 0 \) to \( v \) (initial velocity is 0), and the right side integrates from \( 0 \) to \( x \):
\[ \int_0^v v \, dv = \int_0^x \frac{q}{m}(4 - 3x) \, dx \]

### Step 5: Perform the Integrations
The left side becomes:
\[ \frac{v^2}{2} \bigg|_0^v = \frac{v^2}{2} \]
The right side becomes:
\[ \frac{q}{m} \left( 4x - \frac{3x^2}{2} \right) \bigg|_0^x = \frac{q}{m} \left( 4x - \frac{3x^2}{2} \right) \]

### Step 6: Set the Two Sides Equal
Setting the two sides equal gives:
\[ \frac{v^2}{2} = \frac{q}{m} \left( 4x - \frac{3x^2}{2} \right) \]

### Step 7: Determine When the Particle Comes to Rest Again
The particle comes to rest again when \( v = 0 \). Therefore, we set the equation to find when:
\[ 0 = 4x - \frac{3x^2}{2} \]
Rearranging gives:
\[ 3x^2 - 8x = 0 \]
Factoring out \( x \):
\[ x(3x - 8) = 0 \]
This gives us two solutions:
1. \( x = 0 \)
2. \( 3x - 8 = 0 \Rightarrow x = \frac{8}{3} \)

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