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**Two electrons are fired at each other....**

## Homework Statement

Two electrons are fired at 3.5 x 10

^{6}m/s directly at each other. Calculate the smallest possible distance between the two electrons.

## Homework Equations

Ee = k(q1)(q2) / r

Ek = (0.5)mv

^{2}

## The Attempt at a Solution

I looked at the problem a different way. I though of one electron as stationary, while the other electron moved towards the first at 7.0 x 10

^{6}m/s (twice the speed of the single electron).

I let r represent the minimum distance between the particles. The work required to bring an electron from a distance of infinity to distance r is

k(1.6 x 10

^{-19})(1.6 x 10

^{-19}) / r

(Charge on one electron is 1.6 x 10

^{-19}C)

(k is 9 x 10

^{9})

The electron's original kinetic energy is converted into electrical potential energy. At the minimum distance apart, the kinetic energy is zero, so all the kinetic energy has been converted into electrical potential energy.

Thus,

(0.5)(9.11 x 10

^{-31})(7.0 x 10

^{6})

^{2}= k(1.6 x 10

^{-19})(1.6 x 10

^{-19}) / r

(mass of electron is 9.11 x 10

^{-31}kg)

Which means r = 1.0 x 10

^{-11}m

However, this answer is wrong. The actual answer is 4.5 x 10

^{-6}m. I can't figure out what I did wrong.

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