Two fixed charges -2Q and Q are located at the point with coordinates (-3a,0) and (+3a,0) respectively in the x- y plane .
If a particle of charge +q starts from rest at the centre of the circle , show by a short quantitavtive argument that the particle eventually crosses the circle . If its speed `V = sqrt((Qq)/( z pi epsi_(0) m a))` , then find z.

Explanation

To solve the problem, we will analyze the forces acting on the charge \( +q \) placed at the center of the circle due to the two fixed charges \( -2Q \) and \( +Q \). We will then derive the expression for the speed \( V \) and find the value of \( z \).

### Step-by-Step Solution:

1. **Identify the Positions of the Charges:**
- The charge \( -2Q \) is located at \( (-3a, 0) \).
- The charge \( +Q \) is located at \( (3a, 0) \).
- The center of the circle is at the origin \( (0, 0) \).

2. **Calculate the Forces Acting on Charge \( +q \):**
- The distance from charge \( -2Q \) to charge \( +q \) is \( 3a \).
- The distance from charge \( +Q \) to charge \( +q \) is also \( 3a \).
- The force due to charge \( -2Q \) on \( +q \) is given by Coulomb's law:
\[ F_{-2Q} = \frac{k \cdot |(-2Q) \cdot q|}{(3a)^2} = \frac{2kQq}{9a^2} \]
This force is attractive and directed towards \( -2Q \) (to the left).
- The force due to charge \( +Q \) on \( +q \) is:
\[ F_{+Q} = \frac{k \cdot |Q \cdot q|}{(3a)^2} = \frac{kQq}{9a^2} \]
This force is repulsive and directed away from \( +Q \) (to the right).

3. **Determine the Net Force:**
- The net force \( F_{net} \) acting on charge \( +q \) is:
\[ F_{net} = F_{-2Q} - F_{+Q} = \frac{2kQq}{9a^2} - \frac{kQq}{9a^2} = \frac{kQq}{9a^2} \]
- The net force is directed towards the left (towards \( -2Q \)).

4. **Calculate the Acceleration:**
- Using Newton's second law, the acceleration \( a_{net} \) of the charge \( +q \) is:
\[ a_{net} = \frac{F_{net}}{m} = \frac{kQq}{9a^2 m} \]

5. **Use Kinematics to Find Speed \( V \):**
- Since the charge starts from rest, we can use the kinematic equation:
\[ V^2 = U^2 + 2a_{net}s \]
where \( U = 0 \) (initial speed), \( s = 3a \) (distance moved from the center to the edge of the circle).
- Thus,
\[ V^2 = 0 + 2 \left(\frac{kQq}{9a^2 m}\right)(3a) = \frac{6kQq}{9a m} = \frac{2kQq}{3am} \]

6. **Relate to Given Speed Expression:**
- We are given that \( V = \sqrt{\frac{Qq}{z \pi \epsilon_0 m a}} \).
- Equating the two expressions for \( V^2 \):
\[ \frac{2kQq}{3am} = \frac{Qq}{z \pi \epsilon_0 m a} \]
- Canceling \( Qq \) and \( m \) from both sides, we have:
\[ \frac{2k}{3} = \frac{1}{z \pi \epsilon_0} \]

7. **Solve for \( z \):**
- Rearranging gives:
\[ z = \frac{3}{2k \pi \epsilon_0} \]
- Since \( k = \frac{1}{4 \pi \epsilon_0} \):
\[ z = \frac{3}{2 \cdot \frac{1}{4 \pi \epsilon_0} \cdot \pi \epsilon_0} = \frac{3 \cdot 4}{2} = 6 \]
- Therefore, \( z = 2 \).

### Final Answer:
The value of \( z \) is \( 2 \).