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, we need to analyze the motion of the charged particle in the electric field. Here's a step-by-step breakdown of the solution:

### Step 1: Understanding the Initial Conditions
The charge \( q \) of mass \( m \) enters the electric field at the origin with an initial velocity given by:
\[ \vec{v}_0 = v_0 \hat{i} + v_0 \hat{j} \]
This means that the charge has equal components of velocity in both the x and y directions.

### Step 2: Analyzing the Electric Field
The electric field is given by:
\[ \vec{E} = -E_0 \hat{j} \]
This indicates that the electric field is directed downward (in the negative y-direction). The force on the charge due to the electric field is given by:
\[ \vec{F} = q \vec{E} = -q E_0 \hat{j} \]
This force will cause an acceleration in the negative y-direction.

### Step 3: Finding the Acceleration
Using Newton's second law, the acceleration \( \vec{a} \) of the charge can be calculated as:
\[ \vec{a} = \frac{\vec{F}}{m} = \frac{-q E_0}{m} \hat{j} \]
This shows that the charge experiences a constant acceleration downward.

### Step 4: Motion in the x-direction
In the x-direction, there is no force acting on the charge, so it moves with a constant velocity. The x-component of the velocity remains:
\[ v_x = v_0 \]
The x-coordinate as a function of time \( t \) is given by:
\[ x(t) = v_0 t \]

### Step 5: Motion in the y-direction
In the y-direction, the charge is subject to constant acceleration. The equation of motion for the y-coordinate \( y(t) \) is:
\[ y(t) = v_{0y} t + \frac{1}{2} a_y t^2 \]
Where \( v_{0y} = v_0 \) and \( a_y = -\frac{q E_0}{m} \). Thus, we can write:
\[ y(t) = v_0 t - \frac{1}{2} \frac{q E_0}{m} t^2 \]

### Step 6: Finding the Time to Hit the x-axis
The charge hits the x-axis when \( y(t) = 0 \):
\[ 0 = v_0 t - \frac{1}{2} \frac{q E_0}{m} t^2 \]
Factoring out \( t \):
\[ t \left( v_0 - \frac{1}{2} \frac{q E_0}{m} t \right) = 0 \]
This gives two solutions: \( t = 0 \) (the initial time) and:
\[ t = \frac{2v_0 m}{q E_0} \]

### Step 7: Finding the x-coordinate at this time
Substituting \( t \) back into the equation for \( x(t) \):
\[ x\left(\frac{2v_0 m}{q E_0}\right) = v_0 \left(\frac{2v_0 m}{q E_0}\right) = \frac{2v_0^2 m}{q E_0} \]

### Final Result
Thus, the x-coordinate where the charge hits the x-axis again is:
\[ \boxed{\frac{2v_0^2 m}{q E_0}} \]