Four particles, each of mass m and charge q, are held at the vertices of a square of side 'a'. They are released at t = 0 and move under mutual repulsive forces speed of any particle when its distance from the centre of square doubles is

Options

A. `[(1)/(4pi in_(0))(q^(2))/(ma)(1+(1)/(2sqrt(2)))]^((1)/(2))`
B. `((1)/(4pi in_(0))(q^(2))/(ma))^((1)/(2))`
C. `[(1)/(4pi in_(0))(2q^(2))/(ma^(2))(1+(1)/(2sqrt(2)))]^((1)/(2))`
D. `((1)/(4pi in_(0))(q^(2))/(ma^(2)))^((1)/(2))`

Explanation

To solve the problem of finding the speed of any particle when its distance from the center of the square doubles, we can follow these steps:

### Step 1: Understand the Initial Configuration
We have four particles, each with mass \( m \) and charge \( q \), positioned at the vertices of a square with side length \( a \). The distance from the center of the square to any vertex is given by:
\[ d = \frac{a}{\sqrt{2}} \]

### Step 2: Determine the Initial Potential Energy
The potential energy \( U \) between two charges \( q \) separated by a distance \( r \) is given by:
\[ U = k \frac{q^2}{r} \]
where \( k \) is Coulomb's constant.

For our square configuration:
- There are 4 pairs of charges on the sides, each at a distance \( a \).
- There are 2 pairs of charges on the diagonals, each at a distance \( \sqrt{2}a \).

Thus, the initial potential energy \( U_i \) is:
\[ U_i = 4 \left( k \frac{q^2}{a} \right) + 2 \left( k \frac{q^2}{\sqrt{2}a} \right) = 4k \frac{q^2}{a} + 2k \frac{q^2}{\sqrt{2}a} \]

### Step 3: Determine the Final Configuration
When the particles move away from each other, the distance from the center of the square doubles. The new distance from the center to any vertex becomes:
\[ d' = 2d = 2 \cdot \frac{a}{\sqrt{2}} = \sqrt{2}a \]

### Step 4: Calculate the Final Potential Energy
Now, we need to calculate the final potential energy \( U_f \) when the particles are at this new distance:
- The distance between charges on the sides is now \( 2a \).
- The distance between charges on the diagonals is now \( 2\sqrt{2}a \).

Thus, the final potential energy \( U_f \) is:
\[ U_f = 4 \left( k \frac{q^2}{2a} \right) + 2 \left( k \frac{q^2}{2\sqrt{2}a} \right) = 2k \frac{q^2}{a} + k \frac{q^2}{\sqrt{2}a} \]

### Step 5: Calculate the Change in Potential Energy
The change in potential energy \( \Delta U \) is given by:
\[ \Delta U = U_f - U_i \]

Substituting the expressions for \( U_f \) and \( U_i \):
\[ \Delta U = \left( 2k \frac{q^2}{a} + k \frac{q^2}{\sqrt{2}a} \right) - \left( 4k \frac{q^2}{a} + 2k \frac{q^2}{\sqrt{2}a} \right) \]
\[ = -2k \frac{q^2}{a} - k \frac{q^2}{\sqrt{2}a} \]

### Step 6: Relate Change in Potential Energy to Kinetic Energy
According to the work-energy theorem, the change in kinetic energy \( \Delta K \) is equal to the negative of the change in potential energy:
\[ \Delta K = -\Delta U \]
The initial kinetic energy is zero (since they start from rest), so:
\[ \frac{1}{2} mv^2 = -\Delta U \]

### Step 7: Solve for Speed
Substituting \( \Delta U \) into the equation:
\[ \frac{1}{2} mv^2 = 2k \frac{q^2}{a} + k \frac{q^2}{\sqrt{2}a} \]
\[ v^2 = \frac{4kq^2}{ma} + \frac{2kq^2}{\sqrt{2}ma} \]
\[ v = \sqrt{\frac{4kq^2}{ma} + \frac{2kq^2}{\sqrt{2}ma}} \]

### Final Result
Thus, the speed of any particle when its distance from the center of the square doubles is:
\[ v = \sqrt{\frac{kq^2}{m a} \left(4 + \frac{2}{\sqrt{2}}\right)} \]