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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?

Respuesta :

krlosdark12

Answer:

a) [tex]E=364N/C[/tex]

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:

[tex]x=v_x*t[/tex]

[tex]where:\\x=distance\\v=speed\\t=time[/tex]

so:

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

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

[tex]y=\frac{1}{2}.a.t^2\\\\where:\\y=vertical\_distance\\a=acceleration\\t=time[/tex]

so:

[tex]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[/tex]

now we can use the relation:

[tex]F=m.a=E.q\\so\\E=\frac{m.a}{q}[/tex]

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

Now we can calculate the electric field:

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

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.

The magnitude of the electric field is 364 N/C if the electron misses the upper plate.

The time taken by the electron to travel through plates,

[tex]t = \dfrac {v_x}x[/tex]

Where,

[tex]v_x[/tex]- velocity = [tex]1.6 \times 10^8\; \rm m/s[/tex]

[tex]x[/tex]- distance = 0.2 m

Put the values in the formula,

[tex]t = \dfrac {0.02 }{1.6 \times 10^8}\\\\t = 1.25 \times 10^{-8}[/tex]

Now, acceleration of the electron in verticle direction,

[tex]a = \dfrac {2 \times y }{t^2}[/tex]

Where,

y - verticle distance

So,

[tex]a = \dfrac {2 \times (\frac {0.01 }2 )}{(1.25 \times 10^-8)^2 }\\\\a = 6.4 \times 10^1^3 \rm \; m/s^2[/tex]

The electrical field,

[tex]E = \dfrac {m\times a}{q}[/tex]

Where,

[tex]m[/tex] - mass of electron = [tex]9.1\times 10^{-31} \rm \; kg[/tex]

[tex]q[/tex] - charge = [tex]1.6 \times 10^{-19}\rm \; C[/tex]

So,

[tex]E = \dfrac {9.1\times 10^{-31} \rm \; kg \times 6.4 \times 10^1^3 \rm \; m/s^2}{1.6 \times 10^{-19}\rm \; C}\\\\E = 364 \rm \ N/C \\[/tex]

Therefore, the magnitude of the electric field is 364 N/C.

To know more about the electric field,

https://brainly.com/question/12757739

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