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.

Explanation

To solve the problem of a charged particle of mass \( m \) and charge \( q \) released from rest in a uniform electric field, we need to derive the relationship between the velocity \( v \) of the particle and the distance \( x \) it travels.

### Step-by-Step Solution:

1. **Identify the Force on the Charged Particle**:
The force \( F \) acting on the charged particle in an electric field \( E \) is given by:
\[ F = qE \]

2. **Apply Newton's Second Law**:
According to Newton's second law, the force is also equal to the mass times acceleration:
\[ F = ma \]
Therefore, we can equate the two expressions for force:
\[ ma = qE \]
From this, we can express the acceleration \( a \) as:
\[ a = \frac{qE}{m} \]

3. **Relate Acceleration to Velocity and Distance**:
We know that acceleration can also be expressed as the derivative of velocity with respect to time:
\[ a = \frac{dv}{dt} \]
We can also express \( a \) in terms of velocity and distance:
\[ a = v \frac{dv}{dx} \]
Setting these equal gives us:
\[ v \frac{dv}{dx} = \frac{qE}{m} \]

4. **Separate Variables and Integrate**:
Rearranging the equation, we have:
\[ v \, dv = \frac{qE}{m} \, dx \]
Now, we can integrate both sides. The left side integrates from \( 0 \) to \( v \) and the right side from \( 0 \) to \( x \):
\[ \int_0^v v \, dv = \int_0^x \frac{qE}{m} \, dx \]
The integral of \( v \) gives:
\[ \frac{v^2}{2} \bigg|_0^v = \frac{qE}{m} x \bigg|_0^x \]
This results in:
\[ \frac{v^2}{2} = \frac{qE}{m} x \]

5. **Rearranging the Equation**:
Multiplying both sides by 2 gives:
\[ v^2 = \frac{2qE}{m} x \]

6. **Identifying the Graph**:
The equation \( v^2 = \frac{2qE}{m} x \) is in the form of \( y = kx \), where \( y = v^2 \) and \( k = \frac{2qE}{m} \). This indicates that the graph of \( v^2 \) against \( x \) is a straight line, and since \( v \) is the square root of \( v^2 \), the graph of \( v \) against \( x \) will be a parabola opening to the right.

### Conclusion:
The graph between velocity \( v \) and distance \( x \) will be a parabola symmetric about the x-axis.

---