A positive point charge (q) is projected along (+x) axis electric field is directed along (-y) axis `E = E_0(-hatj)` Find trajectory of charge after x = d.

Options

A. `y = qE_0 d/(m(v_0)^2)*(d/2 -x)`
B. `y = qfE_0d/(m(v_0)^2)*(d/2 +x)`
C. `y = qE_0d/(2m(v_0)^2)*(d/2 -x)`
D. `y = qE_0d/(2m(v_0)^2)*(d/2 +x)`

Answer

To solve the problem, we need to find the trajectory of a positive point charge \( q \) projected along the \( +x \) axis in a uniform electric field directed along the \( -y \) axis. Let's break down the solution step by step.

### Step 1: Understand the motion in the x-direction
The charge is projected along the \( +x \) axis with an initial speed \( v \). Since there is no force acting in the \( x \) direction, the acceleration in the \( x \) direction is zero. Thus, the position of the charge in the \( x \) direction as a function of time \( t \) is given by:
\[ x = vt \]

### Step 2: Determine the force and acceleration in the y-direction
The electric field \( \mathbf{E} \) is directed along the \( -y \) axis, which means the force \( \mathbf{F} \) acting on the charge is:
\[ \mathbf{F} = q\mathbf{E} = -qE_0 \hat{j} \]
The acceleration \( a_y \) in the \( y \) direction can be calculated using Newton's second law:
\[ a_y = \frac{F}{m} = \frac{-qE_0}{m} \]

### Step 3: Write the equation for motion in the y-direction
The initial velocity in the \( y \) direction is zero (since the charge is projected only along the \( x \) axis). The displacement in the \( y \) direction as a function of time \( t \) is given by:
\[ y = v_{y0}t + \frac{1}{2} a_y t^2 = 0 + \frac{1}{2} \left(-\frac{qE_0}{m}\right) t^2 = -\frac{qE_0}{2m} t^2 \]

### Step 4: Relate time \( t \) to the position \( x \)
From the equation \( x = vt \), we can express time \( t \) in terms of \( x \):
\[ t = \frac{x}{v} \]

### Step 5: Substitute \( t \) into the equation for \( y \)
Now, substitute \( t = \frac{x}{v} \) into the equation for \( y \):
\[ y = -\frac{qE_0}{2m} \left(\frac{x}{v}\right)^2 \]
This simplifies to:
\[ y = -\frac{qE_0}{2mv^2} x^2 \]

### Step 6: Find the trajectory when \( x = d \)
To find the trajectory specifically when \( x = d \):
\[ y = -\frac{qE_0}{2mv^2} d^2 \]

### Final Result
Thus, the trajectory of the charge after \( x = d \) is given by:
\[ y = -\frac{qE_0}{2mv^2} d^2 \]