A charge q of mass m enters a region of electric field `E=-E_(0)hatj` with a velocity `v_(0)hati+v_(0)hatj` at origin. Find the x coordinate where the charge hits the x axis again. (Neglect effect of gravity)

Options

A. `(v_(0)^(2)m)/(E_(0)q)`
B. `(2v_(0)^(2)m)/(E_(0)q)`
C. `(v_(0)^(2)m)/(2E_(0)q)`
D. `(3v_(0)^(2)m)/(2E_(0)q)`

Explanation

To solve the problem step by step, we will analyze the motion of the charge in the electric field and apply the equations of motion.

### Step 1: Understand the Given Information
- A charge \( q \) of mass \( m \) enters a region with an electric field \( \mathbf{E} = -E_0 \hat{j} \).
- The initial velocity of the charge is \( \mathbf{v_0} = v_0 \hat{i} + v_0 \hat{j} \).
- The charge starts at the origin (0, 0).

### Step 2: Determine the Forces Acting on the Charge
The electric field exerts a force on the charge given by:
\[ \mathbf{F} = q \mathbf{E} = q (-E_0 \hat{j}) = -q E_0 \hat{j} \]
This force causes an acceleration in the \( y \)-direction.

### Step 3: Calculate the Acceleration
Using Newton's second law, the acceleration \( \mathbf{a} \) is given by:
\[ \mathbf{a} = \frac{\mathbf{F}}{m} = \frac{-q E_0 \hat{j}}{m} = -\frac{q E_0}{m} \hat{j} \]
So, the acceleration in the \( y \)-direction is:
\[ a_y = -\frac{q E_0}{m} \]
And there is no acceleration in the \( x \)-direction:
\[ a_x = 0 \]

### Step 4: Analyze Motion in the \( y \)-Direction
Using the second equation of motion for the \( y \)-direction:
\[ s_y = ut + \frac{1}{2} a_y t^2 \]
Where:
- \( s_y = 0 \) (the charge returns to the x-axis),
- \( u = v_0 \) (initial velocity in the \( y \)-direction),
- \( a_y = -\frac{q E_0}{m} \).

Substituting these values:
\[ 0 = v_0 t + \frac{1}{2} \left(-\frac{q E_0}{m}\right) t^2 \]
This simplifies to:
\[ 0 = v_0 t - \frac{q E_0}{2m} t^2 \]
Factoring out \( t \):
\[ t \left(v_0 - \frac{q E_0}{2m} t\right) = 0 \]
This gives us two solutions:
1. \( t = 0 \) (the initial position),
2. \( t = \frac{2m v_0}{q E_0} \) (the time when the charge hits the x-axis again).

### Step 5: Analyze Motion in the \( x \)-Direction
Since there is no acceleration in the \( x \)-direction, the distance traveled in the \( x \)-direction is given by:
\[ s_x = v_x t \]
Where \( v_x = v_0 \) (initial velocity in the \( x \)-direction). Substituting for \( t \):
\[ s_x = v_0 \left(\frac{2m v_0}{q E_0}\right) \]
This simplifies to:
\[ s_x = \frac{2m v_0^2}{q E_0} \]

### Step 6: Conclusion
The \( x \)-coordinate where the charge hits the x-axis again is:
\[ \boxed{\frac{2m v_0^2}{q E_0}} \]