Question

proton is released from rest inside a region of constant, uniform electric field E1 pointing due North. 34.8 seconds after it is released, the electric field instantaneously changes to a constant, uniform electric field E2 pointing due South. 9.33 seconds after the field changes, the proton has returned to its starting point. What is the ratio of the magnitude of E2 to the magnitude of E1? You may neglect the effects of gravity on the proton.

Answer 1

for electric field E1

force acting F1y = +E1*q

acceleration a1y = F1/m = E1*q/m = E1*(1.6*10^-19/(1.67*10^-27) = E1*9.58*10^7

displacement y1 = voy*t1 + (1/2)*a1*t1^2

velocity after t1 time v1y = voy + a1*t1

for electric field E2

force acting F2 = -E2*q

acceleration a2 = F2/m = -E2*q/m = -E2*9.58*10^7

displacement y2 = v1y*t2 + (1/2)*a2*t2^2

total displacement = 0

y1 + y2 = 0

initial velocity voy = 0

t1 - 34.8 s

t2 = 9.33 s

0 + (1/2)*E1*9.58*10^7*34.8^2 + (E1*9.58*10^7*34.8*9.33) - (1/2)*E2*9.58*10^7*9.33^2 = 0

E2/E1 = 21.4 <<<<---------ANSWER