m = mass of proton
q = charge of proton
During first E1
F = q E1 north
a = q E1/m north
vfinal = (q/m) E1 (14.8)
x final = 0 + 0 +.5 (q/m)E1 (14.8)^2 = (q/m)E1 (110)
Now during E2
x initial = 110 (q/m)E1
Vinitial = (q/m)E1 (14.8)
x final = 0
so
0 = 110(q/m)E1 + 14.8(q/m)E1(8.07)-.5(q/m)E2(8.07^2)
so
.5 E2(8.07)^2 = 110 E1 + 14.8(8.07)E1
32.6 E2 = 110 E1+119 E1 = 229 E1
E2/E1 = 7.04
A proton is released from rest inside a region of constant, uniform electric field E1 pointing due North. 14.8 seconds after it is released, the electric field instantaneously changes to a constant, uniform electric field E2 pointing due South. 8.07 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.
I found that the ratio is just a2/a1. I am not getting the right answer.
1 answer
1 answer