In a uniform electric field if a charge is fired in a direction different from the line of electric field then the trajectory of the charge will be a

Options

A. Straight line
B. Circle
C. Parabola
D. Ellipse

Answer

To determine the trajectory of a charge moving in a uniform electric field at an angle to the field lines, we can break down the problem step by step.

### Step 1: Understand the Setup
We have a uniform electric field \( E \) directed along the x-axis. A charge \( q \) with mass \( m \) is fired with an initial velocity \( v \) at an angle \( \theta \) to the x-axis.

### Step 2: Break Down the Initial Velocity
The initial velocity \( v \) can be resolved into two components:
- The x-component (along the electric field):
\[ u_x = v \cos \theta \]
- The y-component (perpendicular to the electric field):
\[ u_y = v \sin \theta \]

### Step 3: Determine the Accelerations
The charge experiences an acceleration due to the electric field:
- The acceleration in the x-direction (due to the electric field):
\[ a_x = \frac{qE}{m} \]
- The acceleration in the y-direction is zero because there are no forces acting in that direction:
\[ a_y = 0 \]

### Step 4: Write the Equations of Motion
Using the equations of motion, we can express the position of the charge in both the x and y directions as functions of time \( t \):
- For the x-direction:
\[ x = u_x t + \frac{1}{2} a_x t^2 \]
Substituting for \( u_x \) and \( a_x \):
\[ x = (v \cos \theta) t + \frac{1}{2} \left(\frac{qE}{m}\right) t^2 \]

- For the y-direction:
\[ y = u_y t + \frac{1}{2} a_y t^2 \]
Since \( a_y = 0 \):
\[ y = (v \sin \theta) t \]

### Step 5: Eliminate Time to Find the Trajectory
To find the relationship between \( x \) and \( y \), we need to eliminate \( t \) from the equations. From the y-equation, we can express \( t \):
\[ t = \frac{y}{v \sin \theta} \]
Substituting this expression for \( t \) into the x-equation:
\[ x = (v \cos \theta) \left(\frac{y}{v \sin \theta}\right) + \frac{1}{2} \left(\frac{qE}{m}\right) \left(\frac{y}{v \sin \theta}\right)^2 \]
Simplifying this gives:
\[ x = \frac{\cos \theta}{\sin \theta} y + \frac{1}{2} \left(\frac{qE}{m}\right) \frac{y^2}{(v^2 \sin^2 \theta)} \]
This is a quadratic equation in \( y \), which can be expressed as:
\[ y = a x + b y^2 \]
where \( a \) and \( b \) are constants derived from the coefficients.

### Conclusion
The resulting equation indicates that the trajectory of the charge is a **parabola**.