A charge particle of mass m and charge q is released from rest in uniform electric field. Its graph between velocity (v) and distance travelled (x) will be :

Options

A.
B.
C.
D.

Answer

To solve the problem of how the velocity \( v \) of a charged particle changes with the distance \( x \) traveled in a uniform electric field, we can follow these steps:

### Step-by-Step Solution

1. **Understanding the Forces**:
A charged particle of mass \( m \) and charge \( q \) experiences a force \( F \) due to the electric field \( E \). According to Newton's second law, this force can be expressed as:
\[ F = ma \]
where \( a \) is the acceleration of the particle.

2. **Relating Force to Electric Field**:
The force acting on the particle due to the electric field is given by:
\[ F = qE \]
By equating the two expressions for force, we have:
\[ ma = qE \]

3. **Finding Acceleration**:
From the equation \( ma = qE \), we can express the acceleration \( a \) as:
\[ a = \frac{qE}{m} \]

4. **Using Kinematic Relations**:
We know that acceleration can also be expressed in terms of velocity and distance:
\[ a = v \frac{dv}{dx} \]
Setting the two expressions for acceleration equal gives us:
\[ v \frac{dv}{dx} = \frac{qE}{m} \]

5. **Separating Variables**:
Rearranging the equation, we can separate the variables:
\[ v \, dv = \frac{qE}{m} \, dx \]

6. **Integrating Both Sides**:
Now we integrate both sides. The left side integrates to:
\[ \int v \, dv = \frac{v^2}{2} \]
The right side integrates to:
\[ \int \frac{qE}{m} \, dx = \frac{qE}{m} x \]
Thus, we have:
\[ \frac{v^2}{2} = \frac{qE}{m} x + C \]
Since the particle is released from rest, when \( x = 0 \), \( v = 0 \), which gives \( C = 0 \).

7. **Final Equation**:
Therefore, the equation simplifies to:
\[ v^2 = \frac{2qE}{m} x \]

8. **Identifying the Graph**:
The equation \( v^2 = \frac{2qE}{m} x \) represents a parabolic relationship between \( v \) and \( x \). This means that if we plot \( v \) on the y-axis and \( x \) on the x-axis, the graph will be a parabola opening to the right.

### Conclusion
The graph between velocity \( v \) and distance \( x \) for a charged particle in a uniform electric field is a parabola.