A positively charged ring is in y - z plane with its centre at origin . A positive test charge `q_(e )` , held at origin is released along x axis, then its speed

Options

A. Increases continuously
B. Decreases continuously
C. first increases then decreases
D. first decreases then increases

Answer

To solve the problem of a positively charged ring in the y-z plane with a positive test charge \( q_e \) released from the origin along the x-axis, we need to analyze the electric field generated by the ring and how it affects the test charge.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- The charged ring is located in the y-z plane, centered at the origin.
- A positive test charge \( q_e \) is placed at the origin and is released to move along the x-axis.

2. **Electric Field Due to the Ring**:
- The electric field \( \vec{E} \) at a point along the x-axis due to a charged ring can be calculated. The electric field at a distance \( x \) from the center of the ring (along the x-axis) is given by:
\[ E(x) = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q \cdot x}{(x^2 + R^2)^{3/2}} \]
where \( Q \) is the total charge on the ring, \( R \) is the radius of the ring, and \( \epsilon_0 \) is the permittivity of free space.

3. **Direction of the Electric Field**:
- Since both the ring and the test charge are positively charged, the electric field produced by the ring at the location of the test charge will point away from the ring. Thus, when the test charge is released, it will experience a force in the positive x-direction.

4. **Acceleration of the Test Charge**:
- The force \( F \) on the test charge \( q_e \) due to the electric field is given by:
\[ F = q_e \cdot E(x) \]
- The acceleration \( a \) of the test charge can be calculated using Newton's second law:
\[ a = \frac{F}{m} = \frac{q_e \cdot E(x)}{m} \]
where \( m \) is the mass of the test charge.

5. **Finding the Speed of the Test Charge**:
- As the test charge moves along the x-axis, it will accelerate due to the electric field. The speed \( v \) of the test charge can be found using the kinematic equation:
\[ v^2 = u^2 + 2a s \]
where \( u \) is the initial speed (which is 0 since it starts from rest), \( a \) is the acceleration, and \( s \) is the distance traveled along the x-axis.

6. **Conclusion**:
- As the test charge moves away from the origin, it will gain speed due to the electric field created by the positively charged ring. The speed will increase as the test charge moves further along the x-axis.