A proton is released from rest inside a region of constant, uniform electric field E1 pointing due North. 24.8 seconds after it is released, the electric field instantaneously changes to a constant, uniform electric field E2 pointing due South. 8.91 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.
Solution:
Using the relation,
F = qE, and F = ma, so qE = ma and a = qE/m and E = am/q.
In the first part, q = 1.60 * 10ˉ19 C; E = E₁; m = 1.67 * 10ˉ-27 kg, so
a1 = (1.60 * 10-19 C)(E1) / (1.67 *10-27 kg)
= (E1)(9.58 * 107) m/s2.
After 24.8 s, v = a1t = ((E1)(9.58 *
107) m/s²)(24.8 s)
= (E1)(2.38 × 109) m/s.
x (position) = xo + vot + 1/2a1t2
= 0 m + (0 m/s)(24.8 s) + 1/2(E1)(9.58 * 107) (24.3 s)2
= (E1)(2.887 * 1010) m.
In the second part, 8.91 s after the field reversal, the proton has traveled (E2)2.887 * 1010 m back to the initial point.
x = 0 = xo + vot + 1/2a1t2
= (E1)(2.887 * 1010) m +
(E1)(2.38 * 109)(8.91 s) m/s -
1/2a2(8.91 s)2.
So, a2 = ((E1)(2.887 * 1010 m) +
((E2)(2.12 * 1010) m) * 2 / (8.91
s)2
= (E1)(1.262 * 109) m/s².
Since the acceleration is proportional to the electric field
strength, the ratio of the accelerations gives the ratio of the
field strengths:
a2/a1 = E2/E1 =
((E1)(1.262 * 109) m/s2) /
(E1)(9.58 * 107) m/s2
= 13.2.
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