An electron is projected wit velocity `vec(v) = v_(0) hat(x)` in an electric field `vec(E) = E_(0) hat(y)`. Trace the path followed by the electron :-

Options

A. Parabola
B. Circle
C. Straight line in +y direction
D. Straight line in -y direction

Answer

To solve the problem of tracing the path followed by an electron projected in an electric field, we can break down the analysis into several steps:

### Step 1: Understand the Initial Conditions
- The electron is projected with an initial velocity \( \vec{v} = v_0 \hat{x} \).
- The electric field is directed as \( \vec{E} = E_0 \hat{y} \).

**Hint:** Identify the direction of the velocity and the electric field to understand how they interact.

### Step 2: Determine the Force on the Electron
- Since the electron is negatively charged, the force acting on it due to the electric field is given by:
\[ \vec{F} = -e \vec{E} = -e E_0 \hat{y} \]
where \( e \) is the magnitude of the charge of the electron.

**Hint:** Remember that the force on a charged particle in an electric field is in the direction opposite to the field for negative charges.

### Step 3: Analyze Motion Along the X-Axis
- Along the x-axis, there is no force acting on the electron, so it moves with constant velocity:
\[ x(t) = v_0 t \]
- The acceleration in the x-direction \( a_x = 0 \).

**Hint:** For constant velocity motion, the distance is simply the product of velocity and time.

### Step 4: Analyze Motion Along the Y-Axis
- The force acting on the electron causes it to accelerate in the y-direction:
\[ F_y = m a_y \Rightarrow a_y = \frac{-e E_0}{m} \]
- The motion in the y-direction is uniformly accelerated motion starting from rest:
\[ y(t) = u_y t + \frac{1}{2} a_y t^2 \]
where \( u_y = 0 \) (initial velocity in y-direction).

**Hint:** Use the kinematic equation for uniformly accelerated motion to find the position in the y-direction.

### Step 5: Substitute Time in Terms of x
- From the x-axis motion, we can express time \( t \) in terms of \( x \):
\[ t = \frac{x}{v_0} \]
- Substitute this expression for \( t \) into the equation for \( y(t) \):
\[ y = \frac{1}{2} a_y t^2 = \frac{1}{2} \left( \frac{-e E_0}{m} \right) \left( \frac{x}{v_0} \right)^2 \]
Simplifying gives:
\[ y = -\frac{e E_0}{2 m v_0^2} x^2 \]

**Hint:** When substituting, ensure to keep track of the signs and units.

### Step 6: Identify the Path
- The equation \( y = k x^2 \) (where \( k = -\frac{e E_0}{2 m v_0^2} \)) represents a parabola opening downwards.

**Hint:** Recognize the form of the equation to identify the type of curve traced by the motion.

### Conclusion
The path followed by the electron is a parabola.

### Final Answer
The path followed by the electron is a parabolic trajectory.