# Thread: Motion of Charged Particle Question

1. ## Motion of Charged Particle Question

"Two electrons are held, at rest, 1.0x10^-12m apart, then released. With what kinetic energy and speed is each moving when they are a "large" distance apart."

I need help solving for the speed.

I solved for the kinetic energy doing:

/\V = kq/r
/\V = (9.0x10^9 Nm^2/C^2)(1.602x10^-19C)/(1.0x10^-12m)
/\V = 1441.8V

W = /\Ek
/\Ek = q/\V
/\Ek = (1.602X10^-19C)(1441.8V)
/\Ek = 2.3 x 10^16 J

How do I solve for the speed from this point? (answer in the book gives 1.6x10^7m/s)

2. Originally Posted by Jeavus
"Two electrons are held, at rest, 1.0x10^-12m apart, then released. With what kinetic energy and speed is each moving when they are a "large" distance apart."

I need help solving for the speed.

I solved for the kinetic energy doing:

/\V = kq/r
/\V = (9.0x10^9 Nm^2/C^2)(1.602x10^-19C)/(1.0x10^-12m)
/\V = 1441.8V

W = /\Ek
/\Ek = q/\V
/\Ek = (1.602X10^-19C)(1441.8V)
/\Ek = 2.3 x 10^16 J

How do I solve for the speed from this point? (answer in the book gives 1.6x10^7m/s)
Note that the electric force is conservative, so there are no losses to the total mechanical energy: energy is conserved.

So the initial electrical potential energy (not to be confused with the electric potential) will be equal to the final kinetic energy.
$\displaystyle \frac{ke^2}{1.0 \times 10^{-12}~m} = \frac{1}{2}mv^2$

I get $\displaystyle v = 2.249 \times 10^7~m/s$.

Note: This speed is right on the edge of what I would consider using Special Relativity for. (About a tenth of the speed of light.) Please make sure that your instructor does not want you to use this. (If (s)he does, let me know and I'll show you how to alter the calculation.)

-Dan