Question

7.(16 pts) Two parallel plates (see figure below) have a seperation of 5 mm and a potential difference of 1000 V between them. The bottom plate has higher potential. An electron (me = 9.1e-31 kg, ge-_1.6e-19 C) įs released from rest from top plate. (a) Compare the electrical force with gravitational force applied on the electron. (b) Find the speed of the electron when it hits the lower plate, i.e., find the impact speed. (c) Find the impact speed of the electron when two plates are seprated by 2m. (d) Find the impact speed of the electron when two plates are seperated by 2microns. (e) Does the impact speed depend on the value of sepration? What factors determine the impace speed? 8) 0V 1000V

Answer 1

V = er v = F/q r F_e = vq/r F_g = mg Given v = 1000 v m = 9.11 times 10^-31 kg m = 9.11 times 10^-31 kg q = -1.6 times 10^-19 c r = 5 mm = 5 times 10^-3 m electric force F_e = vq/r = 1000 times - 1.6 times 10^-19/5 times 10^-3 = -3.2 times 10^-14 N gravitational force F_e = mg = 9.11 times 10^-31 times 9.81 = 89.36 times 10^-31 N electric force is greater than gravitational forces F_e > F_g distance h = 5mm almostequalto r = 5 times 10^-3 m P.E = KE mgh = 1/2 mv^2 v = squareroot 2gh V = squareroot 2 times 9 times 9 times 10^-3 = squareroot 2 times 9.81 times 10^-3 times 5 = 0.3132 m/s c) When h = 2m V = squareroot 2 times 9.81 times 2 = squareroot 39.24 = 6.26 m/s When h = 2 micron = 2 times 10^-6 m