Small identical balls with equal charges of magnitude 'q' each are fixed at the vertices of a regular 20019-gon (a polygon of 2019 sides) with side `'a'=4 `. At a certain instant, one of the balls is released and a sufficiently long time interval later, the ball adjacent to the first released ball is freed. The kinetic energies of the released balls are found to differ by `K=9xx10^(9) J` at a sufficiently large distance from the polygon. determine the charge q in mC.

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the System
We have a regular polygon with 2019 vertices, each vertex having a charge \( q \). When one charge is released, it will experience a force due to the other charges in the polygon. The potential energy of the system will change when another charge is released.

### Step 2: Initial Energy Calculation
Initially, when the first charge is released, its kinetic energy \( K_1 \) is zero because it starts from rest. The potential energy \( U \) of the system can be calculated using the formula for the potential energy between point charges:
\[ U = k \cdot \frac{q^2}{r} \]
where \( k \) is Coulomb's constant (\( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \)), and \( r \) is the distance between the charges.

### Step 3: Final Energy Calculation
When the first charge is released and moves far away, its potential energy becomes zero, and all the energy is converted into kinetic energy:
\[ E_{\text{final}} = K_1 \]
Thus, we have:
\[ K_1 = U \]

### Step 4: Second Charge Release
When the second charge (adjacent to the first) is released, the potential energy of the system changes because one less charge is contributing to the potential energy. The new potential energy \( U' \) will be:
\[ U' = U - k \cdot \frac{q^2}{a} \]
where \( a \) is the distance between the two adjacent charges.

### Step 5: Kinetic Energy of Second Charge
The kinetic energy \( K_2 \) of the second charge when it is released is:
\[ K_2 = U' = U - k \cdot \frac{q^2}{a} \]

### Step 6: Energy Difference
The difference in kinetic energies of the two released charges is given as \( K = K_1 - K_2 \):
\[ K = U - (U - k \cdot \frac{q^2}{a}) = k \cdot \frac{q^2}{a} \]

### Step 7: Substitute Known Values
We know that \( K = 9 \times 10^9 \, \text{J} \) and \( a = 4 \, \text{m} \):
\[ 9 \times 10^9 = k \cdot \frac{q^2}{4} \]
Substituting \( k = 9 \times 10^9 \):
\[ 9 \times 10^9 = 9 \times 10^9 \cdot \frac{q^2}{4} \]

### Step 8: Solve for \( q^2 \)
Dividing both sides by \( 9 \times 10^9 \):
\[ 1 = \frac{q^2}{4} \]
Thus,
\[ q^2 = 4 \]

### Step 9: Calculate \( q \)
Taking the square root:
\[ q = \sqrt{4} = 2 \, \text{C} \]

### Step 10: Convert to mC
Since the question asks for the charge in millicoulombs (mC):
\[ q = 2 \, \text{C} = 2000 \, \text{mC} \]

### Final Answer
The charge \( q \) is \( 2000 \, \text{mC} \).

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