A positively charged particle starts at rest 25cm from a second positively charged particle which is held stationary throughout the experiment. The first particle is released and accelerates directly

Question

A positively charged particle starts at rest 25cm from a second positively charged particle which is held stationary throughout the experiment. The first particle is released and accelerates directly away from the second particle. When the first particle has moved 25cm, it has reached a velocity of 2 ​ m / s . What is the maximum velocity (in m / s ) that the first particle will reach ?

TestHub · Accepted Answer

Energy from release (25 cm) to the given point (50 cm): $\frac{1}{2} m (2)^{2} = k q_{1} q_{2} (\frac{1}{0.25} - \frac{1}{0.5}) = 2 k q_{1} q_{2}$ (SI units).

At infinity all of the initial potential energy is kinetic: $\frac{1}{2} m v_{ma x}^{2} = \frac{k q _{1} q _{2}}{0.25} = 4 k q_{1} q_{2}$ .

Dividing the two: $\frac{v _{ma x}^{2}}{2} = \frac{4}{2}$ ⇒ $v_{ma x} = 2$ m/s.

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