Question

Two parallel plates 1.0 cm apart are equally and oppositely charged. An electron is released from rest at the surface of the negative plate and simultaneously a proton is released from rest at the surface of the positive plate. How far from the negative plate is the point at which the electron and proton pass each other?

Answer 1

The distance of the negative plate at which the electron and proton pass each other is; 0.9995

Particle Acceleration

The formula for the particle acceleration is;

a = qE/m

Where;

q is charge

E is electric field

m is mass

Now, we will make d to denote the distance as from the negative plate.

Thus, the travel distance by the proton at that same time will be 1 - d.

Thus, from Newton's equation of motion we can say that;

d = ½a_e*t²

1 - d = ½a_p*t²

Where a_e and a_p are acceleration of electron and proton respectively.

Dividing both equations gives us;

d/(1 - d) = a_e/a_p

This can be rewritten as;

d/(1 - d) = m_p/m_e

Where;

m_e is mass of electron = 9.1 × 10^(-31) kg

m_p is mass of proton = 1.67 × 10^(-27) kg

Thus;

d/(1 - d) = (1.67 × 10^(-27))/(9.1 × 10^(-31))

d/(1 - d) = 1835

d = 1835 - 1835d

1836d = 1835

d = 0.9995 m

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