A proton is projected in the positive x direction into aregion of uniform electric field E = -6.60 multiplied by 105i N/C att = 0. The proton travels 7.40 cm as it comes to rest. (a) Formula for calculating the value of the acceleration (a) of theproton is
a = Eq / m
= -(-6.60 multiplied by 105i N/C)(1.6*10-19C) /(1.67*10-27kg)
= -6.32*1013m/s2i
So magnitude of the acceleration of the particle is a =6.32*1013m/s2
And its direction is alongnegative x direction
(b) Distance (S) traveled =7.40cm = 0.074m
Then we know the formula v2 - u2 = 2aS
0 - u2 = 2aS
= 2(-6.32*1013m/s2)(0.0740m)
u = 3.1*106m/s
So initial speed (u) of the proton is u = 3.1*106m/s
(c) Now time taken is t = u / a
= (3.1*106m/s) /(6.32*1013m/s2)
= 0.484*10-7s