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?
2007-09-29 13:35:02 · 4 answers · asked by biscuits 2 in Science & Mathematics ➔ Physics
We need to make the assumption that the plates are infinite and that the field between them is uniform.
The electron and the proton are subjected to the same force intensity f. Integrate twice with regards to time, position x = 1/2 f/m*t^2.
The particles will cross when :
xelectron = 1/2 f/melectron * t^2 = 1cm - 1/2 f/mproton * t^2
xelectron = 1cm - melectron/mproton * xelectron
xelectron = 1/(1*melectron/mproton) cm
xelectron = 0.999456 cm
Considering there is only a 1 digit precision in the distance between the plates the exact result does not mean much. We can only say that the electron almost came all the way while the proton nearly didn't move at all.
2007-09-29 14:24:03 · answer #1 · answered by twilight1138 2 · 0⤊ 0⤋
Let the distance the electron ravels be s. The the distance the proton travels is .01 - s (keep it in MKS units) The force acting on both of them is the same (since thsy both have a unit charge) but their accellerations will be different since the proton is..... (Fµck!! I don't remember..... This gettin' old sucks) Anyway, look up the rest mass of the proton and electron and calculate the acceleration of each (in terms of F). Then
s = (1/2)a(electron)t² and
.01-s = (1/2)a(proton)t²
Solve them for s and you're done (t is a constant, it's the time it takes for them to get to the point they pass each other)
Doug
2007-09-29 14:13:43 · answer #2 · answered by doug_donaghue 7 · 0⤊ 0⤋
In a circuit containing a parallel-plate capacitor and a skill source (including a battery), the skill source grants potential for costs to visit the plates of the capacitor - detrimental costs on one plate and valuable ones on the different plate. This effectively creates an electric powered field between the plates of the capacitor. This field shops potential and that's the potential that flows around the area between the plates - no longer the expenses. The pass of potential between the plates effectively keeps the continuity of 'present day' between the capacior plates.
2016-10-20 08:21:57 · answer #3 · answered by ? 4 · 0⤊ 0⤋
tricky situation. look into using yahoo or google. that may help!
2014-10-31 21:09:37 · answer #4 · answered by william 3 · 0⤊ 2⤋