Question

An electron and a proton are held on an x axis, with the electron at x = + 1.000 m

and the proton at x = - 1.000 m . If a second electron is initially at x = + 20.00 m on the x axis and is given an initial velocity of 450 m/s towards the origin, will it reach and origin? And how close will it get to the origin if it doesn't?

Answer 1

Answer:

  r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

Explanation:

For this exercise we must use conservation of energy

the electric potential energy is

          U = $k \frac{q_1q_2}{r_{12}}$

for the proton at x = -1 m

          U₁ =$- k \frac{e^2 }{r+1}$

for the electron at x = 1 m

          U₂ = $k \frac{e^2 }{r-1}$

starting point.

        Em₀ = K + U₁ + U₂

        Em₀ = $\frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1}$

final point

         Em_f = $k e^2 ( -\frac{1}{r_2 +1} + \frac{1}{r_2 -1})$

   

energy is conserved

        Em₀ = Em_f

        \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e^2 (- \frac{1}{r_2 +1} + \frac{1}{r_2 -1})              

       

        \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e²(  $\frac{2}{(r_2+1)(r_2-1)}$ )

we substitute the values

½ 9.1 10⁻³¹ 450 + 9 10⁹ (1.6 10⁻¹⁹)² [ $- \frac{1}{20+1} + \frac{1}{20-1}$ ) = 9 109 (1.6 10-19) ²( $\frac{2}{r_2^2 -1}$ )

          2.0475 10⁻²⁸ + 2.304 10⁻³⁷ (5.0125 10⁻³) = 4.608 10⁻³⁷ ( $\frac{1}{r_2^2 -1}$ )

          2.0475 10⁻²⁸ + 1.1549 10⁻³⁹ = 4.608 10⁻³⁷     $\frac{1}{r_2^2 -1}$

          $\frac{2.0475 \ 10^{-28} }{1.1549 \ 10^{-37} } = \frac{1}{r_2^2 -1}$

          r₂² -1 = (4.443 10⁸)⁻¹

           

          r2 = $\sqrt{1 + 2.25 10^{-9}}$

          r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m