Question

A proton gun fires a proton from midway between two plates, A and B, which are separated by a distance of 10.9 cm; the proton initially moves at a speed of 101 km/s toward plate B. Plate A is kept at zero potential, and plate B at a potential of 617 V. With what speed will it hit plate A?

Answer 1

given

d = 10.9 m = 0.109 m

VA = 0

VB = 617 volts

vi = 101 km/s

vf = ?

here th

workdone on proton = change in kinetic energy of proton

F.d = (1/2)*m*(vf^2 - vi^2)

F*d*cos(180) = (1/2)*m*(vf^2 - vi^2)

-F*d = (1/2)*m*(vf^2 - vi^2)

-q*E*d = (1/2)*m*(vf^2 - vi^2)

-q*(VB - VA) = (1/2)*m*(vf^2 - vi^2)

vf^2 - vi^2 = -2*q*(VB - VA)/m

vf = sqrt(vi^2 - 2*q*(VB - VA)/m)

= sqrt( (101*10^3)^2 - 2*1.6*10^-19*(617 - 0)/(1.67*10^-27))

= inderminant

so, the proton can not hit the plate B.