A uniform electric field of magnitude 640 N/C exists between two parallel plates

Question

A uniform electric field of magnitude 640 N/C exists between two parallel plates that are 4.40 cm apart. A proton is released from the positive plate at the same instant that an electron is released from the negative plate. (a) Determine the distance from the positive plate at which the two pass each other. (Ignore the electrical attraction between the proton and electron.) Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully.

Explanation / Answer

a)

F = q E = 1.6e-19 * 640 = 1.024e-16 N

ap = F/mp = 1.024e-16/1.673e-27 = 6.1207e10 m/s2

ae = F/me = 1.024e-16/9.109e-31 = 1.12416e14 m/s2

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xp = 0.5 ap t2 = 0.5 * 6.1207e10 * t2 = 3.0604e10 t2

xe = o.5 ae t2 = 0.5 * 1.12416e14 * t2 = 0.56208e14 t2

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xp + xe = 4.4 cm = 4.4e-2 m

3.0604e10 t2 + 0.56208e14 t2 = 4.4e-2

5.6239e14 t2 = 4.4e-2

t = 8.8452e-9 s

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xp = 0.5 ap t2 = 0.5 * 6.1207e10 * (8.8452e-9*8.8452e-9) = 2.39 m

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b)

mp = m_Na = 22.9898 u = 22.9898*1.6605e-27 = 3.8175e-26 kg

me = m_Cl = 35.453 u = 35.453*1.6605e-27 = 5.8870e-26 kg

F = q E = 1.6e-19 * 640 = 1.024e-16 N

ap = F/mp = 1.024e-16/3.8175e-26 = 2.6824e9 m/s2

ae = F/me = 1.024e-16/5.8870e-26 = 1.7394e9 m/s2

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xp = 0.5 ap t2 = 0.5 * 2.6824e9 * t2 = 1.3412e9 t2

xe = o.5 ae t2 = 0.5 * 1.7394e9 * t2 = 0.8697e9 t2

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xp + xe = 4.4 cm = 4.4e-2 m

1.3412e9 t2 + 0.8697e9 t2 = 4.4e-2

2.2109e9 t2 = 4.4e-2

t = 4.4611e-6 s

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xp = 0.5 ap t2 = 0.5 * 2.6824e9 * (4.4611e-6*4.4611e-6) = 2.67 cm

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