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: Determine the acceleration of the particle
The force acting on the particle due to the electric field is given by:
\[ \vec{F} = q \vec{E} = q E_0 \hat{j} \]
Using Newton's second law, the acceleration \( \vec{A} \) of the particle can be calculated as:
\[ \vec{A} = \frac{\vec{F}}{m} = \frac{q E_0 \hat{j}}{m} \]
Thus, the acceleration in the \( y \)-direction is:
\[ A_y = \frac{q E_0}{m} \]

### Step 2: Calculate the velocity of the particle over time
Since the particle is released from rest, its initial velocity is zero. The velocity \( v_y \) in the \( y \)-direction at any time \( t \) can be calculated using the formula:
\[ v_y = A_y \cdot t = \left(\frac{q E_0}{m}\right) t \]

### Step 3: Determine the position of the particle over time
The position of the particle in the \( y \)-direction as a function of time can be found using the equation of motion:
\[ y(t) = \frac{1}{2} A_y t^2 = \frac{1}{2} \left(\frac{q E_0}{m}\right) t^2 \]
The \( x \)-coordinate remains constant at \( x_0 \).

### Step 4: Calculate the angular momentum about the origin
The angular momentum \( L \) of the particle about the origin is given by:
\[ L = r_{\perpendicular} \cdot p \]
where \( r_{\perpendicular} \) is the perpendicular distance from the line of action of the momentum to the origin, and \( p \) is the momentum of the particle.

At any time \( t \):
- The perpendicular distance \( r_{\perpendicular} \) is equal to \( x_0 \).
- The momentum \( p \) is given by:
\[ p = m v_y = m \left(\frac{q E_0}{m}\right) t = q E_0 t \]

Thus, the angular momentum can be expressed as:
\[ L = x_0 \cdot (q E_0 t) \]
This simplifies to:
\[ L = q E_0 x_0 t \]

### Conclusion
The angular momentum of the particle about the origin is:
\[ L = q E_0 x_0 t \]
This indicates that the angular momentum is directly proportional to time \( t \).

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