Question

Situation: two electrons, each with mass m and charge q, are released from positions very far from each other. With respect to a certain reference frame, electron A has initial nonzero speed v toward electron B in the positive x direction, and electron B has initial speed 3v toward electron A in the negative x direction. The electrons move directly toward each other along the x axis (very hard to do with real electrons). As the electrons approach each other, they slow due to their electric repulsion. This repulsion eventually pushes them away from each other.Question: What is the minimum separation r_min that the electrons reach?Express your answer in term of q, m, v, and k (where k=).r_min =

Answer 1

Answer:

$r_{min}=\frac{kq^2}{5m_ev^2}$

Explanation:

The total kinetic energy of both electrons will be electrostatic potential energy, when the electrons reach the minima distance due to electrostatic repulsion. Then, you have:

$E_{k}=U_E\\\\\frac{1}{2}m_ev_1^2+\frac{1}{2}m_ev_2^2=k\frac{q^2}{r_{min}}$

me: mass of the electron

q: charge of the electron

k: Coulomb's constant

you take into account that v2=3v1=3v and do rmin the subject of the formula:

$\frac{1}{2}m_e[v^2+9v^2]=5m_ev^2=k\frac{q^2}{r_{min}}\\\\r_{min}=\frac{kq^2}{5m_ev^2}$