Question

An electron is ejected with speed v=107 m/s into the uniform field created by the flat and parallel sheets of the figure. The field is directed vertically downwards and is null except in the space between the sheets. The electron It enters through a point located at an equal distance between the two sheets. Yes when you leave field, the electron passes just through the edge of the sheet: a) Find the intensity of the field b) Find the direction of the speed of the electron when it leaves the field.

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Answer to a math question An electron is ejected with speed v=107 m/s into the uniform field created by the flat and parallel sheets of the figure. The field is directed vertically downwards and is null except in the space between the sheets. The electron It enters through a point located at an equal distance between the two sheets. Yes when you leave field, the electron passes just through the edge of the sheet: a) Find the intensity of the field b) Find the direction of the speed of the electron when it leaves the field.

Expert avatar
Neal
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45 Answers
Para resolver este problema, utilizaremos la ley de la fuerza eléctrica para determinar la intensidad del campo y la dirección de la velocidad del electrón cuando sale del campo.

a) Para hallar la intensidad del campo, utilizamos la siguiente fórmula:

F = q \cdot E

donde F es la fuerza eléctrica, q es la carga del electrón y E es la intensidad del campo.

Sabemos que la fuerza eléctrica es la fuerza centrípeta que actúa sobre el electrón, por lo que podemos escribir:

F = \frac{{m \cdot v^2}}{{r}}

donde m es la masa del electrón, v es su velocidad y r es el radio de la trayectoria del electrón.

El radio de la trayectoria del electrón es la distancia entre las dos láminas, que llamaremos d. Dado que el electrón entra por un punto situado a igual distancia entre las láminas, podemos decir que r = d/2.

Reemplazando estos valores en la ecuación de la fuerza eléctrica, obtenemos:

\frac{{m \cdot v^2}}{{r}} = q \cdot E

\frac{{m \cdot v^2}}{{d/2}} = q \cdot E

Resolviendo para E, obtenemos:

E = \frac{{2 \cdot m \cdot v^2}}{{q \cdot d}}

Por lo tanto, la intensidad del campo es:

E = \frac{{2 \cdot m \cdot v^2}}{{q \cdot d}}

b) Para hallar la dirección de la velocidad del electrón cuando sale del campo, podemos utilizar la ley de conservación de la energía cinética:

\frac{{1}}{{2}} \cdot m \cdot v^2 = q \cdot V

donde V es el potencial eléctrico en el borde de la lámina.

La energía cinética inicial del electrón es igual a su energía cinética final más la energía potencial eléctrica ganada:

\frac{{1}}{{2}} \cdot m \cdot v^2 = q \cdot V

Dado que el electrón pasa justamente por el borde de la lámina cuando sale del campo, el potencial eléctrico en ese punto es cero (V = 0). Por lo tanto, la velocidad del electrón cuando sale del campo es la misma que su velocidad inicial, v = 107 m/s.

Por lo tanto, la dirección de la velocidad del electrón cuando sale del campo es la misma que su dirección inicial, que es vertical hacia abajo.

Answer:
a) La intensidad del campo es E = \frac{{2 \cdot m \cdot v^2}}{{q \cdot d}}
b) La dirección de la velocidad del electrón cuando sale del campo es vertical hacia abajo.

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