A positive point charge (q) is projected along (+x) axis with some velocity from origin, electric field is directed along (y) axis `E = E_0(hatj)` Find trajectory of charge in form of x.

Options

A. Option1 `y = qE_0 x^2/(m(v_0)^2)`
B. Option2 `y = qE_0 2x/(m(v_0)^2)*`
C. Option3 `y = qE_0x^2/(2m(v_0)^2)`
D. Option4 `y = qE_0x^3/(2m(v_0)^2)`

Answer

To solve the problem of finding the trajectory of a positive point charge projected along the x-axis in an electric field directed along the y-axis, we can follow these steps:

### Step 1: Understand the motion in the x-direction
Since the charge is projected along the positive x-axis with an initial velocity \( v_0 \) and there is no electric field acting in the x-direction, the motion in the x-direction is uniform. Therefore, the position of the charge in the x-direction can be expressed as:
\[ x(t) = v_0 t \]

### Step 2: Analyze the motion in the y-direction
The electric field \( E \) is directed along the y-axis, given as \( E = E_0 \hat{j} \). The force \( F \) acting on the charge \( q \) due to this electric field can be calculated using:
\[ F = qE = qE_0 \]
This force will cause an acceleration \( a_y \) in the y-direction, which can be calculated using Newton's second law:
\[ a_y = \frac{F}{m} = \frac{qE_0}{m} \]

### Step 3: Write the equation of motion in the y-direction
Since the initial velocity in the y-direction is zero (the charge is only projected in the x-direction), we can use the kinematic equation for displacement in the y-direction:
\[ y(t) = u_y t + \frac{1}{2} a_y t^2 \]
Substituting \( u_y = 0 \) and \( a_y = \frac{qE_0}{m} \), we get:
\[ y(t) = 0 + \frac{1}{2} \left( \frac{qE_0}{m} \right) t^2 = \frac{qE_0}{2m} t^2 \]

### Step 4: Substitute for time \( t \)
From the equation for \( x(t) \), we can express time \( t \) in terms of \( x \):
\[ t = \frac{x}{v_0} \]
Now, substitute this expression for \( t \) into the equation for \( y(t) \):
\[ y = \frac{qE_0}{2m} \left( \frac{x}{v_0} \right)^2 \]
This simplifies to:
\[ y = \frac{qE_0}{2m v_0^2} x^2 \]

### Step 5: Final trajectory equation
Thus, the trajectory of the charge in the form of \( y \) as a function of \( x \) is:
\[ y = k x^2 \]
where \( k = \frac{qE_0}{2m v_0^2} \).

### Summary
The trajectory of the charge projected along the x-axis in the presence of an electric field directed along the y-axis is a parabolic path described by the equation:
\[ y = \frac{qE_0}{2m v_0^2} x^2 \]