Answer to Question #258870 in Electricity and Magnetism for K-Lo

Question

In a particle accelerator, an electron is fired horizontally into an electric field between 2 charged plates at a speed of 2.0 (106) m/s.  It enters the field halfway between the 2 plates that are separated by 40 mm.  The electric field has a strength of 3000 N/C.  Determine the distance the electron will travel horizontally before it hits one of the plates.

Answer 1

Let us determine the acceleration of the electron perpendicular to its initial motion. The electric force is $F = eE$ and the acceleration is $a = \dfrac{eE}{m}.$ The electron is initially at the distance $s = d/2$ from the plate, let us calculate the time interval needed to move at such a distance.

$s = \dfrac{at^2}{2}\,, \;\;\; t = \sqrt{\dfrac{2s}{a}} = \sqrt{\dfrac{d}{a}} = \sqrt{\dfrac{md}{eE}}$ .

$t = \sqrt{\dfrac{9.1\cdot10^{-31}\,\mathrm{kg}\cdot 4\cdot10^{-2}\,\mathrm{m}}{1.6\cdot10^{-19}\,\mathrm{C}\cdot3\cdot10^3\,\mathrm{N/C}}} = 8.7\cdot10^{-9}\,\mathrm{s}.$

The distance travelled is $vt = 2\cdot10^6\,\mathrm{m/s}\cdot 8.7\cdot10^{-9}\,\mathrm{s} = 0.0174\,\mathrm{m}.$