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

Answer

To solve the problem, we need to find the speed of any particle when its distance from the center of the square doubles after being released. Here’s a step-by-step solution:

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

### Step 2: Determine the Final Distance
We want to find the speed of the particles when their distance from the center of the square doubles:
\[ d_{\text{final}} = 2 \cdot d_{\text{initial}} = 2 \cdot \frac{a}{\sqrt{2}} = \sqrt{2}a \]

### Step 3: Relate the Side Length of the Square to the Distance
When the distance from the center doubles, the side length of the square also changes. Let the new side length be \( a' \). The new distance from the center to a vertex is:
\[ d_{\text{final}} = \frac{a'}{\sqrt{2}} \]
Setting this equal to our final distance, we have:
\[ \frac{a'}{\sqrt{2}} = \sqrt{2}a \implies a' = 2a \]

### Step 4: Calculate Initial Potential Energy
The initial potential energy \( U_i \) of the system can be calculated by considering the potential energy between each pair of charges. There are 6 pairs of charges:
- 4 pairs of adjacent charges (distance \( a \))
- 2 pairs of diagonal charges (distance \( a\sqrt{2} \))

The potential energy for adjacent pairs is:
\[ U_{\text{adjacent}} = 4 \cdot \frac{kq^2}{a} \]
The potential energy for diagonal pairs is:
\[ U_{\text{diagonal}} = 2 \cdot \frac{kq^2}{a\sqrt{2}} = \frac{\sqrt{2}kq^2}{a} \]
Thus, the total initial potential energy is:
\[ U_i = 4 \cdot \frac{kq^2}{a} + \frac{\sqrt{2}kq^2}{a} = \left(4 + \frac{\sqrt{2}}{1}\right) \frac{kq^2}{a} \]

### Step 5: Calculate Final Potential Energy
After the particles move to the new positions, the potential energy \( U_f \) can be calculated similarly:
\[ U_f = 4 \cdot \frac{kq^2}{2a} + 2 \cdot \frac{kq^2}{2a\sqrt{2}} = 2 \cdot \frac{kq^2}{a} + \frac{kq^2}{\sqrt{2}a} \]
Thus, the total final potential energy is:
\[ U_f = \left(2 + \frac{1}{\sqrt{2}}\right) \frac{kq^2}{a} \]

### Step 6: Apply Conservation of Energy
According to the conservation of energy:
\[ U_i = U_f + K \]
Where \( K \) is the total kinetic energy at the final state. Since all particles have the same speed \( v_0 \):
\[ K = 4 \cdot \frac{1}{2} mv_0^2 = 2mv_0^2 \]

### Step 7: Set Up the Equation
Substituting the expressions for potential energy into the conservation of energy equation:
\[ \left(4 + \frac{\sqrt{2}}{1}\right) \frac{kq^2}{a} = \left(2 + \frac{1}{\sqrt{2}}\right) \frac{kq^2}{a} + 2mv_0^2 \]

### Step 8: Solve for \( v_0 \)
Rearranging gives:
\[ 2mv_0^2 = \left(4 + \frac{\sqrt{2}}{1} - 2 - \frac{1}{\sqrt{2}}\right) \frac{kq^2}{a} \]
\[ 2mv_0^2 = \left(2 + \frac{\sqrt{2} - 1/\sqrt{2}}{1}\right) \frac{kq^2}{a} \]
\[ v_0^2 = \frac{\left(2 + \frac{\sqrt{2} - 1/\sqrt{2}}{1}\right) kq^2}{2ma} \]
Taking the square root gives:
\[ v_0 = \sqrt{\frac{\left(2 + \frac{\sqrt{2} - 1/\sqrt{2}}{1}\right) kq^2}{2ma}} \]

### Final Answer
The speed of any particle when its distance from the center of the square doubles is:
\[ v_0 = \sqrt{\frac{(4 + \sqrt{2}) kq^2}{4ma}} \]