Question

Two capacitor plates are equally and oppositely charged. They are separated by 1.5cm . 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 the positive plate be at x = 0 and the negative plate be at x0 = 0.015m. Since the plates are equally and oppositely charged, there is a uniform electric field between them = E + S/e0 where S = surface charge density on the positive plate and e0 = 8.85x10^-12 C^2/(N-m^2). The proton experiences a force Fp

Fp = mp*ap = qE --> ap = qE/mp mp = mass of proton

and the electron feels a force Fe = me*ae = -qE --> ae = -qE/me me = mass of electron

Now they both move according to x = 1/2 at^2 + x1 where x1 = starting postion and we want to find the time t when they are both at the same place call it x = x2 so

1/2ap*t^2 = 1/2 ae*t^2 + x1 --> 2x1 = (ap -ae)*t^2 --> t = sqrt(2x1/(ap-ae))

Now substitute into equation of motion for proton:

x = x2 = 1/2 ap*t^2 = 1/2 ap*2x1/(ap-ae) = x1*(ap/(ap - ae))

Now do some messy algebra

x2 = x1*(qE/mp/(qE/mp + qE/me)) note you can cancel the qE out!

x2 = x1*( 1/mp/(1/mp+1/me)) = x1*(me/(me+mp)) Now mp ~ 1800 me so

x2 = x1*(1/(1801)) = 0.015/1801 m = 83.3x10^-6 m from positive plate