Question

A proton is projected in the positive x direction into aregion of uniform electric field E = -6.60 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 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


Does anyone see where I went wrong? All help is appreciated!

Answer 1

Part a

The magnitude of the acceleration of proton particle is .

Part b

The initial velocity of the moving proton particle is .

Part c

The time taken by the proton particle is .