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 when it is placed at the center of the circle formed by the two fixed charges -2Q and Q. We will also derive the expression for the speed of the charge +q and find the value of z.

### Step-by-Step Solution:

1. **Identify the Configuration**:
- We have two fixed charges: -2Q at (-3a, 0) and Q at (3a, 0).
- The charge +q starts from rest at the center of the circle, which is at the origin (0, 0).

2. **Calculate the Forces**:
- The distance from the charge +q to -2Q is 3a, and the distance from +q to Q is also 3a.
- The force due to charge -2Q on +q is given by Coulomb's law:
\[ F_{-2Q} = k \frac{|-2Q \cdot q|}{(3a)^2} = k \frac{2Qq}{9a^2} \]
- The force due to charge Q on +q is:
\[ F_{Q} = k \frac{|Q \cdot q|}{(3a)^2} = k \frac{Qq}{9a^2} \]

3. **Net Force Calculation**:
- The net force acting on +q is the vector sum of the forces due to -2Q and Q. Since -2Q is to the left and Q is to the right, we have:
\[ F_{net} = F_{-2Q} - F_{Q} = k \frac{2Qq}{9a^2} - k \frac{Qq}{9a^2} = k \frac{(2Q - Q)q}{9a^2} = k \frac{Qq}{9a^2} \]

4. **Determine Acceleration**:
- Using Newton's second law, \( F = ma \):
\[ a = \frac{F_{net}}{m} = \frac{k \frac{Qq}{9a^2}}{m} = \frac{kQq}{9ma^2} \]

5. **Use Kinematic Equation**:
- Since the charge starts from rest, we can use the kinematic equation:
\[ v^2 = u^2 + 2as \]
- Here, \( u = 0 \) (initial velocity), \( a = \frac{kQq}{9ma^2} \), and \( s = 3a \):
\[ v^2 = 0 + 2 \left(\frac{kQq}{9ma^2}\right)(3a) = \frac{6kQq}{9m} = \frac{2kQq}{3m} \]

6. **Express Speed**:
- Taking the square root gives:
\[ v = \sqrt{\frac{2kQq}{3m}} \]

7. **Compare with Given Expression**:
- We are given that \( v = \sqrt{\frac{Qq}{z \pi \epsilon_0 m a}} \).
- We know that \( k = \frac{1}{4 \pi \epsilon_0} \), so substituting \( k \) into our expression:
\[ v = \sqrt{\frac{2 \cdot \frac{1}{4 \pi \epsilon_0} Qq}{3m}} = \sqrt{\frac{Qq}{6 \pi \epsilon_0 m}} \]

8. **Finding z**:
- Now we can equate the two expressions for \( v \):
\[ \sqrt{\frac{Qq}{6 \pi \epsilon_0 m}} = \sqrt{\frac{Qq}{z \pi \epsilon_0 m}} \]
- This implies:
\[ z = 6 \]

### Final Result:
The value of \( z \) is **6**.