Aproton is released from rest inside a region of constant, uniform electric field e 1 pointing due north. 27.3 s after it is released, the electric field instantaneously changes to a constant, uniform electric field e 2 pointing due south. 3.03 s after the field changes, the proton has returned to its starting point. what is the ratio of the magnitude of e 2 to the magnitude of e 1 ? you may neglect the effects of gravity on the proton.

E2/E1 =99.2

Explanation:

time after release of E1 (t) = 27.3 s

time after release of E2 (t') = 3.03 s

acceleration (a) =

where

after 27.3 s

velocity (V) = a x t = x 27.3  =

distance to turning point (s) = 0.5a = 0.5 x x =

now for its return back to its starting point

acceleration (a') =

total distance S' = distance to turning point + distance from turning point to starting point

S' = S + vt' + 0.5 a'

S' = + ( x 3.03) + (0.5 x x )

S' is the distance at the starting point and = 0

0 =

multiplying both side by M/Q we have

4.59.E2 = 372.65E1 + 82.72E1

4.59.E2 = 455.37E1

E2/E1 = 455.37 / 4.59

E2/E1 =99.2