Question

Two parallel plates, separated by 0.20 m, are connected to a 12-V battery. An electron released from rest at a location 0.10 m from the negative plate. When the electron arrives at a distance 0.050 m from the positive plate, what is the speed of the electron?

Answer 1

Strength of electric field   E   =   V / d

                                                =   12 / 0.20

                                                =   60.0   N/C

   

   Force on the electron   F   =   q * E

                                          =   1.6 * 10-19 * 60.0

                                          =   9.6 * 10-18   N

   Acceleration of electron      a   =   F/m

                                                =   9.6 * 10-18 / 9.1 * 10-31

                                                =   1.055 * 1013   m/s2

   According to third equation of motion

   v2   =   u2   +   2 * a * s

   s   =   distance covered   =   (0.10   -   0.050)

                                       =   0.050   m

   u   =   initial speed   =   0

   =>   v2   =   02   +   2 * 1.055 * 1013 * 0.050

               =   1.055 * 1012

   Final speed of electron   v   =   v1.055 * 1012

                                             =   1.027 * 106   m/s