A pair of oppositely charged parallel plates are separated by a distance of 4.4

Question

A pair of oppositely charged parallel plates are separated by a distance of 4.4 cm with a potential difference of 513 V between the plates. A proton is released from rest at the positive plate at the same time that an electron is released from rest at the negative plate. Disregard any interaction between the proton and the electron. How long does it take for the paths of the proton and the electron to cross? 6.55339 times 10-9 s. How fast will the electron be traveling when the particles' paths cross? 1.34209 times 107 m/s. How fast will the proton be traveling when the particles' paths cross? 7307.27 m/s. How much time will elapse before the proton reaches the opposite plate? 2.80929 times 10-7 s.

Explanation / Answer

Electric field= V/d = 513/ 0.044 = 11659 N/c apprx

Force exeperienced by electron and proton= 1.6 x 10-19 ( 11659) =18654.54 x 10-19 N apprx

a( accelertion of electron = F/m = 18654.54 x 10-19/ 9.1 x 10-31 = 2050 x 1012m/s ^2 apprx

a( accelertaion of proton) = F/m= 18654.54 x 10-19/ 1.67 x 10-27= 11170.4 x 108 m/s^2 apprx

A) let the distance covered be { x } and{ 0.044 -x }

uisng kinematic equations

x = 0.5 (2050 x1012) t2 --------------eq(1)

( 0.044-x) = 0.5( 11170.4 x 108) t2-----------eq(2)

solving eq(1) and (2), we get

t = 6.549 x 10-9 seconds aprx

b)velocity of electron

v = 2050x 1012 x ( 6.549 x 10-9) = 13425.45x 103 m/s = 1.342 x 107 m/s apprx

c) velocity of protion,

v = u+at = 0 + 11170.4 x 108x ( 6.549 x 10-9)= 73154 . 949 x 10-1 m/s = 7315. 5 m/s apprx

d) 0.044 = 0.5 ( 11170.4 x 108) t2

t = 0.0028067 x 10-4= 2.8 x 10^ -7 s apprx

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