The path of a charged particle projected into a uniform transverse electric field is

Options

A. circle
B. hyperbola
C. parabola
D. ellipse

Answer

To solve the question regarding the path of a charged particle projected into a uniform transverse electric field, we can follow these steps:

### Step 1: Understand the Forces Acting on the Charged Particle
When a charged particle is placed in an electric field, it experiences a force due to the electric field. The force \( F \) acting on a charged particle with charge \( Q \) in an electric field \( E \) is given by:
\[ F = Q \cdot E \]
This force acts in the direction of the electric field.

### Step 2: Analyze the Motion of the Charged Particle
If the charged particle is projected into a uniform transverse electric field, it means that the initial velocity of the particle is perpendicular to the direction of the electric field. For example, if the electric field is directed vertically (upwards), the particle is projected horizontally.

### Step 3: Break Down the Motion into Components
The motion of the charged particle can be analyzed in two dimensions:
- The horizontal motion (along the x-axis) where there is no force acting on the particle, so it moves with constant velocity.
- The vertical motion (along the y-axis) where the particle experiences a constant force due to the electric field, causing it to accelerate.

### Step 4: Apply Kinematic Equations
In the horizontal direction (x-axis):
- The displacement \( x \) is given by:
\[ x = V_x \cdot t \]
where \( V_x \) is the initial horizontal velocity and \( t \) is the time.

In the vertical direction (y-axis):
- The displacement \( y \) is given by the equation of motion under constant acceleration:
\[ y = \frac{1}{2} a_y t^2 \]
where \( a_y = \frac{F}{m} = \frac{Q \cdot E}{m} \) is the acceleration due to the electric field.

### Step 5: Combine the Equations
Since \( y \) depends on \( t^2 \) and \( x \) depends linearly on \( t \), we can eliminate \( t \) from these equations. From the horizontal motion, we can express \( t \) as:
\[ t = \frac{x}{V_x} \]
Substituting this into the vertical motion equation gives:
\[ y = \frac{1}{2} a_y \left(\frac{x}{V_x}\right)^2 \]
This equation represents a parabolic relationship between \( y \) and \( x \).

### Conclusion
The path of the charged particle projected into a uniform transverse electric field is a **parabola**.

### Final Answer
The correct option is **parabola**.

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