A charged particle (a proton) is moving rightward between two parallel charged p

Question

A charged particle (a proton) is moving rightward between two parallel charged plates (one on the left and the other on the right, facing each other) separated by distance d = 2.00mm. The plate potentials are V subscript 1 space equals space minus 70.0 space V on the left and V subscript 2 space equals space minus 50.0 V on the right. The particle is slowing from an initial speed of 90.0km/s at the left plate. What is its speed just as it reaches plate 2 on the right, in the unit m/s? The unit charge is 1.6 space x space 10 to the power of minus 19 end exponent C . The mass of proton is 1.67 space x space 10 to the power of minus 27 end exponent k g .

Explanation / Answer

deacceleration = 20*1.6*10-19/1.67*10-27 = 20*108 m/sec2

v initial = 90km/sec = 90000m/sec

by equation

V2=U2-2as

let final velocity be V

u = 90000m/sec , s= 0.002 , a = 20*108

V = 89.95 km/sec

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