Two identical point charges are moving in free space. When they are 60 cm apart, their velocity vectors are equal in modulus and make angles of  ${45}^0$ from the line joining them as shown in the figure. If at this instant, their total kinetic energy is equal to their potential energy, the distance of closest approach between them = _______________.

Distance of Closest Approach Calculation

Two identical point charges are moving in free space. At a separation of 60 cm, their velocity vectors have equal magnitudes and make a 45° angle with the line joining them. The total kinetic energy equals the total potential energy at this instant.

Total Kinetic Energy (K.E.) = 2 × (1/2 m u²)
Total Potential Energy (P.E.) = k q² / r

Using conservation of energy and vector components, the relation for loss in kinetic energy and gain in potential energy leads to the equation:

m u² / 2 = k q² (1/r₁ - 1/r₂)

where:
r₁ = initial separation (60 cm),
r₂ = minimum separation (distance of closest approach)

Solving the relation using momentum and geometry, the minimum separation is found to be:

r₂ = (2/3) × r₁ = (2/3) × 60 cm = 40 cm

However, considering velocity component and energy relations, the calculation gives:

m w = 6 / 0.1
x = 6 cm

Final distance of closest approach = 6 cm