Question

Outbox Sent Junk An electron is trapped between two large parallel charged plates of a capacitive system. The plates are separated by a distance of 1 cm and there is vacuum in the region between the plates. The electron la initially found midway between the plates with a kinetic energy of 14.9 V and with Its velocity directed toward the negative plate. Trash Archive Mailboxes Part A Today How close to the negative plate will the electron get if the potential difference between the plates is 100 ? (1 ov = 1.8 x 10-19 J) O 0.149 cm 0 cm 0.351 cm 0.485 cm 0.202 cm Submit Request Answer < Return to Assignment Provide Feedback 26,476 55 31

Answer 1

Given datar Kinetic energy TE = qv = 14.9ev distance d= 1cm= 1x102 m Potential difference V = 100v lev = 1.6x10-lag mass of electron m=9.1 X10 31 kg 29v speed v V 100 = lou wc 10-2 2x14.9*7.681019 9.1610-31 V=2.28 *106 m/s Electric field E = charge on electron q =-16 = -1.6 x 10 lac Acceleration a = -1.6x10-19 x 104 9.1 *1031 a = -1.76x1015 m/s2 O XE - v²u²=2as S final velocity vso (2.28 <106)? = 2 (-1.76 $1015) S=0.147 cm mid of seperation 2/2 = 0.5cm distance of electron from negative plate -0.5 -0,147 - 0.352 CM