Question

An electron is projected with an initial speed v0 = 1.10 x 10⁶ m/s into the uniform field between the parallel plates. The distance between the plates is 1 cm and the length of the plates is 2 cm. Assume that the field between the plates is uniform and directed vertically downward, and that the field outside the plates is zero. The electron enters the field at a point midway between the plates. E = N/C
(a) If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field.
(b) Suppose that the electron is replaced by a proton with the same initial speed. Would the proton hit one of the plates?

Answer 1

Answer:

a) $E=364N/C$

b) No

Explanation:

A) Because the electron is affected by an acceleration force in this case by the electric field, we can use the formulas of 2-dimension movement.

We will assume the electron missed the upper plate, so we need to calculate the time to travel all the way through the plate, that is:

$x=v_x*t$

$where:\\x=distance\\v=speed\\t=time$

so:

$t=\frac{x}{v_x}=\frac{0.02m}{1.6\cdot 10^6m/s}\\\\t=1.25\cdot10^{-8}$

the electron experiences an accelerated motion in the vertical direction, so we can obtain the acceleration of the electron:

$y=\frac{1}{2}.a.t^2\\\\where:\\y=vertical\_distance\\a=acceleration\\t=time$

so:

$a=\frac{2.y}{t^2}\\\\a=\frac{2*(\frac{0.01}{2}m)}{(1.25\cdot10^{-8}s)^2}\\\\a=6.4\cdot10^{13} m/s^2$

now we can use the relation:

$F=m.a=E.q\\so\\E=\frac{m.a}{q}$

$where:\\\\E=electric\_field\\m=electron\_mass=9.1\cdot10^{-31}kg\\q=Charge=1.6\cdot10^{-19}\\a=acceleration$

Now we can calculate the electric field:

$E=\frac{9.1\cdot10^{-31}kg\cdot6.4\cdot10^{13}m/s^2}{1.6\cdot10^{-19}C}\\\\E=364N/C$

B) Because the proton has the same charge but positive it will go in the opposite direction, so because we assume the electron didn't touch the plate, the proton won't.