re at rest and separated by a distance $ r $. The initial kinetic energy (K.E.) of the system is zero because the particles are not moving. The initial potential energy (P.E.) is given by the electrostatic potential energy formula:

$ \text{Initial P.E.} = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q^2}{r} $

When the particles move closer such that their separation is reduced to $ \frac{r}{2} $, they gain kinetic energy. The final kinetic energy (assuming that both particles gain the same speed $ v $) can be expressed as:

$ \text{Final K.E.} = 2 \times \frac{1}{2} m v^2 = m v^2 $

The final potential energy when the separation is $ \frac{r}{2} $ is:

$ \text{Final P.E.} = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q^2}{r/2} = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{2q^2}{r} $

According to the conservation of energy principle:

$ (\text{Initial K.E.} + \text{Initial P.E.}) = (\text{Final K.E.} + \text{Final P.E.}) $

Substituting the expressions for the initial and final energies:

$ \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q^2}{r} = m v^2 + \frac{1}{4 \pi \varepsilon_0} \cdot \frac{2q^2}{r} $

Rearranging terms to solve for $ m v^2 $:

$ m v^2 = \frac{1}{4 \pi \varepsilon_0}\left(\frac{2q^2}{r} - \frac{q^2}{r}\right) = \frac{q^2}{4 \pi \varepsilon_0} \cdot \frac{1}{r} $

Then:

$ m v^2 = \frac{q^2}{4 \pi \varepsilon_0} \cdot \frac{1}{r} $

Solving for $ v $:

$ v = \frac{q}{\sqrt{4 \pi \varepsilon_0 m r}} $

Thus, the speed of the particles when their separation is reduced to $ \frac{r}{2} $ is given by:

$ v = \frac{q}{\sqrt{4 \pi \varepsilon_0 m r}} $

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