Three identical spheres each of radius a and mass m are placed such that their centers lie on the vertices of an equilateral triangle of side length 3a. The spheres are non conducting and each one carries a charge Q uniformly distributed over its surface. All the three spheres are simultaneously released and after releasing they move under the influence of their mutual electrostatic forces only. Based on this information, answer the following questions :
Suppose that some time after the release, the distance between the spheres becomes 5a and the speed of each sphere becomes v. Then, at this instant, the rate of change of potential energy of the system will be:

Options

A. Zero
B. `-(3sqrt3KQ^(2))/(25a^(2))v
C. `-((KQ^(2))/(3a^(2)))v
D. `-((KQ^(2))/(2a^(2)sqrt3))v

Explanation

To solve the problem, we need to determine the rate of change of potential energy of the system of three identical charged spheres when they are at a distance of 5a from each other and moving with a speed v.

### Step-by-Step Solution:

1. **Understanding the Forces**:
Each sphere carries a charge Q. The electrostatic force \( F \) between any two spheres can be calculated using Coulomb's law:
\[ F = k \frac{Q^2}{r^2} \]
where \( k = \frac{1}{4\pi \epsilon_0} \) and \( r \) is the distance between the centers of the two spheres.

2. **Calculating the Force at Distance 5a**:
When the distance between the spheres is 5a, the force between any two spheres becomes:
\[ F = k \frac{Q^2}{(5a)^2} = \frac{k Q^2}{25a^2} \]

3. **Net Force on Each Sphere**:
Each sphere experiences forces due to the other two spheres. The net force \( F_{net} \) acting on one sphere can be resolved into components. The horizontal components of the forces from the other two spheres will cancel out, and we will only consider the vertical components:
\[ F_{net} = 2F \cos(30^\circ) = 2 \left(\frac{k Q^2}{25a^2}\right) \left(\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3} k Q^2}{25a^2} \]

4. **Using Newton's Second Law**:
According to Newton's second law, the net force is equal to mass times acceleration:
\[ F_{net} = m \cdot a \]
Therefore, we can relate the acceleration \( a \) to the net force:
\[ m \cdot a = \frac{\sqrt{3} k Q^2}{25a^2} \]
This gives us:
\[ a = \frac{\sqrt{3} k Q^2}{25ma^2} \]

5. **Kinetic Energy and Potential Energy**:
The total mechanical energy \( E \) of the system is conserved, which means:
\[ E = U + K \]
where \( U \) is the potential energy and \( K \) is the kinetic energy.

6. **Rate of Change of Potential Energy**:
By differentiating the energy conservation equation with respect to time, we get:
\[ \frac{dU}{dt} + \frac{dK}{dt} = 0 \]
Thus,
\[ \frac{dU}{dt} = -\frac{dK}{dt} \]

7. **Calculating Kinetic Energy**:
The kinetic energy \( K \) of the system is given by:
\[ K = \frac{3}{2} mv^2 \]
Therefore,
\[ \frac{dK}{dt} = 3m v \frac{dv}{dt} \]

8. **Substituting into the Rate of Change of Potential Energy**:
Substituting \( \frac{dK}{dt} \) into our equation gives:
\[ \frac{dU}{dt} = -3mv \frac{dv}{dt} \]

9. **Finding \( \frac{dv}{dt} \)**:
From our earlier equation for acceleration:
\[ \frac{dv}{dt} = \frac{F_{net}}{m} = \frac{\sqrt{3} k Q^2}{25ma^2} \]

10. **Final Expression for Rate of Change of Potential Energy**:
Now substituting \( \frac{dv}{dt} \) into the equation for \( \frac{dU}{dt} \):
\[ \frac{dU}{dt} = -3mv \left(\frac{\sqrt{3} k Q^2}{25ma^2}\right) \]
The mass \( m \) cancels out:
\[ \frac{dU}{dt} = -\frac{3\sqrt{3} k Q^2 v}{25a^2} \]

### Conclusion:
The rate of change of potential energy of the system when the distance between the spheres is 5a and the speed of each sphere is v is:
\[ \frac{dU}{dt} = -\frac{3\sqrt{3} k Q^2 v}{25a^2} \]