Physics, 20.02.2020 18:38 o10889084

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. What is the minimum separation rmin that the electrons reach? Express your answer in term of q, m, v, and k (where k=14πϵ0)

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Answer from: elizabethhubbe

The minimum separation reached by the electrons is $\dfrac{ke^{2}}{4mv^{2}}$ .

The given problem is based on the concept of total energy. The system of two electrons is an isolated system, and the internal forces acting on the system are only conservative forces: therefore, the total energy of the system must be constant.

If we assume the electric potential energy to be zero at the beginning (because the electrons are very far from each other), then the initial energy is just the sum of the kinetic energy of the two electrons. So total energy is given as,

$E_{t}=\dfrac{1}{2}m(v^{2}) +\dfrac{1}{2}m((-3v)^{2})\\\\E_{t}=\dfrac{1}{2}mv^{2} +\dfrac{9}{2}mv^{2}\\\\E_{t}=5mv^{2}$ .................................................................(1)

When the two electrons reach the distance of minimum approach, part of the kinetic energy has been converted into electric potential energy, so the total final energy is:

$E'_{t}=\dfrac{1}{2}mv'^{2} +\dfrac{1}{2}mv'^{2} +\dfrac{k(e)(e)}{r}$ ..................................................................(2)

Here,

v' is the final speed of the two electrons.

e is the charge of the electron.

r is distance of minimum approach.

Applying the conservation of linear momentum for the isolated system as,

$mv+m(-3v)=mv'+mv'\\\\mv-3mv=2mv'\\\\-2mv=2mv'\\\\v'=v$

Substituting into (2) we get,

$E'_{t}=\dfrac{1}{2}mv^{2}+ \dfrac{1}{2}mv^{2}+\dfrac{ke^{2}}{r}\\\\E'_{t}=mv^{2}+\dfrac{ke^{2}}{r}$

Equate equation (1) and (2) to find the distance of minimum approach as,

$5mv^{2}=mv^{2}+\dfrac{ke^{2}}{r}\\\\ r=\dfrac{ke^{2}}{4mv^{2}}$

Thus, we can conclude that the minimum separation reached by the electrons is $\dfrac{ke^{2}}{4mv^{2}}$ .

Learn more about the Conservation of momentum here:

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Answer from: aislamgomes22

Distance of minimum approach: $r=\frac{ke^2}{4mv^2}$

Explanation:

The system of two electrons is an isolated system, and the internal forces acting on the system are only conservative forces: therefore, the total energy of the system must be constant.

The total energy is the sum of the kinetic energy and the electric potential energy.

$E_i = \frac{1}{2}m(v)^2+\frac{1}{2}m(-3v)^2=\frac{1}{2}mv^2+\frac{9}{2}mv^2=5mv^2$ (1)

where v and (-3v) are the initial velocities of the 2 electrons, while m is the mass of each electron.

Later, when the two electrons reach the distance of minimum approach, part of the kinetic energy has been converted into electric potential energy, so the total final energy is:

$E_f=\frac{1}{2}mv'^2+\frac{1}{2}mv'^2+\frac{k(e)(e)}{r}$ (2)

where

v' is the final speed of the two electrons (at the distance of minimum approach, they have the same speed, because the velocity of the center of mass of the system must be zero)

$\frac{k(e)(e)}{r}$ is the final electric potential energy of the system, with

e = charge of the electron

r = distance of minimum approach

Also, the system is isolated, so we can apply the law of conservation of momentum:

$mv+m(-3v)=mv'+mv'\\mv-3mv=2mv'\\-2mv=2mv'\\v'=-v$

Substituting into (2) we get

$E_f=\frac{1}{2}mv^2+\frac{1}{2}mv^+\frac{ke^2}{r}=mv^2+\frac{ke}{r}$

And now we can finally equate the initial and final energy (1)=(2) to find the distance of minimum approach:

$5mv^2=mv^2+\frac{ke^2}{r}\\\rightarrow r=\frac{ke^2}{4mv^2}$

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