A charged particle of mass m and charge q is released from rest the position `(x_0,0)` in a uniform electric field `E_0hatj`. The angular momentum of the particle about origin.

Options

A. is zero
B. is constant
C. increases with time
D. decreases with time

Explanation

To find the angular momentum of a charged particle of mass \( m \) and charge \( q \) released from rest at the position \( (x_0, 0) \) in a uniform electric field \( \vec{E} = E_0 \hat{j} \), we can follow these steps:

### Step 1: Identify the forces acting on the particle
The particle experiences an electric force due to the electric field. The force \( \vec{F} \) acting on the particle is given by:
\[ \vec{F} = q \vec{E} = q E_0 \hat{j} \]

### Step 2: Calculate the acceleration of the particle
Using Newton's second law, the acceleration \( \vec{a} \) of the particle can be calculated as:
\[ \vec{F} = m \vec{a} \implies \vec{a} = \frac{\vec{F}}{m} = \frac{q E_0 \hat{j}}{m} \]

### Step 3: Determine the velocity of the particle at time \( t \)
Since the particle starts from rest, the initial velocity \( u = 0 \). The velocity \( \vec{v} \) at time \( t \) is given by:
\[ \vec{v} = u + \vec{a} t = 0 + \left( \frac{q E_0}{m} \hat{j} \right) t = \frac{q E_0 t}{m} \hat{j} \]

### Step 4: Find the position of the particle at time \( t \)
The position \( \vec{r} \) of the particle at time \( t \) can be found using:
\[ \vec{r} = \vec{r_0} + \vec{v} t = (x_0, 0) + \left( 0, \frac{q E_0 t^2}{2m} \right) = \left( x_0, \frac{q E_0 t^2}{2m} \right) \]

### Step 5: Calculate the angular momentum about the origin
The angular momentum \( \vec{L} \) of a particle about a point is given by:
\[ \vec{L} = \vec{r} \times m \vec{v} \]
Substituting \( \vec{r} \) and \( \vec{v} \):
\[ \vec{L} = \left( x_0, \frac{q E_0 t^2}{2m} \right) \times m \left( 0, \frac{q E_0 t}{m} \right) \]
Calculating the cross product:
\[ \vec{L} = m \left( x_0 \hat{i} + \frac{q E_0 t^2}{2m} \hat{j} \right) \times \left( 0 \hat{i} + \frac{q E_0 t}{m} \hat{j} \right) \]
This results in:
\[ \vec{L} = m \left( x_0 \cdot \frac{q E_0 t}{m} \hat{k} - 0 \right) = q E_0 x_0 t \hat{k} \]

### Final Result
The angular momentum of the particle about the origin is:
\[ \vec{L} = q E_0 x_0 t \hat{k} \]