Question

Two capacitor plates are equally and oppositely charged. They are separated by 1.4cm . An electron is released from rest at the surface of the negative plate and, at the same time, a proton is released from rest at the surface of the positive plate.

Where do the electron and proton pass each other? Give your answer as a distance from the positive plate.

Answer 1

Let d = 1e-2 (m) be the separation of the plates.

The forces on the electron and the proton are the same, but the accelerations and starting points are different.

Electron:
- Force = -eE
- Initial position: x(0) = d
- Initial speed: 0
- Mass: m_e

m_e*x'' = -eE
=> x'' = -eE/m_e
x(t) = x(0) - (eE/m_e)*t^2/2
= d - (eE/(2m_e))*t^2

Proton:
- Force = eE
- Initial position: x(0) = 0
- Initial speed: 0
- Mass: m_p

m_p*x'' = eE
=> x'' = eE/m_p
x(t) = x(0) + (eE/m_p)*t^2/2
= (eE/(2m_p))*t^2

These two meet at the time t when:
(eE/2m_p))*t^2 = x_p(t) = x_e(t) = d - (eE/(2m_e))*t^2

(eE/2)*t^2 * (1/m_p + 1/m_e) = d
t^2 = 2d/(eE) * 1/(1/m_p + 1/m_e)
= 2d/(eE) * m_p*m_e/(m_e + m_p)
= [2*d*m_p*m_e)/((eE)*(m_e+m_p))]

x_p(t) is thus
= (eE/2m_p))*t^2
= (eE/2m_p))*[2*d*m_p*m_e)/((eE)*(m_e+m_p)...
= d*m_e/(m_e+m_p)
= 1.4 (cm) * m_e/(m_e + m_p)
= 1.4 (cm) * 9.1e-31/(9.1e-31 + 1.67e-27)
= 1.4 (cm) * 9.1e-31/(e-27 * (1.67 + 0.00091)
= 1.4 (cm) * 9.1e-31/(1.67e-27)
= 1.4 (cm) * (9.1/1.67)*e-4
= 5.45*e-4 (cm)
= 5.45e-6 (m)
= 5.45 (