Question

A proton is released from rest inside a region of constant, uniform electric field E1 pointing due North. 22.3 seconds after it is released, the electric field instantaneously changes to a constant, uniform electric field E2 pointing due South. 2.61 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

Consider case 1: The kinematics equation is expressed as follows The proton is at rest initially, so u=0 2x 2x Substitute 22.3 s for t in expression a 2x (22.3s) 497.29 The velocity is expressed as follows v=u+alt Substitute 0 for u and 22.3 s for t in expression v = u + at 2x 497.29 2x 22.3

Consider case 2: The kinematics equation is expressed as follows 2x Substitutefor , 2.61 s for t in equation x =-vt +-ar. 22.3 r-2.61s)+-a, (2.61s 22.3 2 12(2.61) 22.3 2.61 Now, the ratio of acceleration is, ai 2r142(2.61) 22.3 6 E1 497.29 90.09

Answer 2

Ratio of acceleration=E2/E1
Since distance travelled is equal.
E2*t22/E1*t12=1
E2/E1=t12/t22=73