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anyanavicka [17]
3 years ago
14

2. A Se lanza un electrón con rapidez inicial v0 = 1.60×106 m/s hacia el interior de un campo uniforme entre las placas paralela

s de la figura E21.33. Suponga que el campo entre las placas es uniforme y está dirigido verticalmente hacia abajo, y que el campo fuera de las placas es igual a cero. El electrón ingresa al campo en un punto equidistante de las dos placas. A) Si el electrón apenas elude la placa superior al salir del campo, encuentre la magnitud del campo eléctrico. B) Suponga que en la figura E21.33 el electrón es sustituido por un protón con la misma rapidez inicial v0. ¿Golpearía el protón una de las placas? Si el protón no golpea ninguna de las placas, ¿cuáles serían la magnitud y la dirección de su desplazamiento vertical, cuando sale de la región entre las placas? C) Compare las trayectorias que recorren el electrón y el protón, y explique las diferencias. D) Analice si es razonable ignorar los efectos de la gravedad para cada partícula.
Physics
1 answer:
Vanyuwa [196]3 years ago
5 0

Answer:

A)     E = 145.6 N / C , B)  y= 2,8 10-7 m with a downward direction

C) he shape of the trajectory of the two particles is to simulate a parabola,

D)     F_{e} /F_{g} = 10³⁴

Explanation:

A) For this exercise we use Newton's second law to find the acceleration of the electron, where the force is electric

           F = m a  

           - e E = m a

          a = - e E / m

with the field directed downward, the acceleration is in the vertical upward direction.

We look for how much the electron moves with kinematics, in the x direction there is no acceleration,

x axis (parallel to plates)

           x = v₀ t

           t = x / v₀

y axis (perpendicular to plates)

          y = y₀ + v_{oy} t + ½ a t²

Let's take the zero of the system in the middle of the plates y₀ = 0, also the initial vertical velocity is zero (v_{oy} = 0) the width of the plate is known

          y = ½ a t²

we substitute

         y = ½ (e E /m)  (x / v₀)²

         y = ½ e x2 /m v₀²   E

we look for the electric field

        E = 2 m y v₀² / e x²

where to use this expression the length and width of the condenser must be known, suppose that the length is x = l = 1 cm = 1 10⁻² m and the width is y = 0.5 mm = 0.5 10⁻³ m

let's calculate

         E = 2  9.1 10⁻³¹ 0.5 10⁻³ (1.6 10⁶)² / (1.6 10⁻¹⁹ (1 10⁻²)²)

         E = 145.6 N / C

B) The electron is exchanged for a proton

Let's look for the vertical displacement, in this case as the proton has a positive charge it moves towards the bottom of the plates

          y = ½ e x² / m v₀² E

          y = ½ 1.6 10⁻¹⁹ 1 10⁻⁴ / (1.67 10⁻²⁷ (1.6 10⁶)²   145.6

          y = 28.4375 10⁻⁸ m

since the distance between the plates is 0.5 10-3 m, the proton passes the condensate because its deflection is very small

In summary, its displacement is y= 2,8 10-7 m and with a downward direction (the same direction of the electric field)

C) The shape of the trajectory of the two particles is to simulate a parabola, but one for having a negative charge (electron) the force is upwards and the other for having a positive charge (proton) the trajectory is downwards

D) The force of gravity

           F_{g} = G m M / R²

electron

          Between the electron and the positive charges of the conducting plate

           F_{g}= 6.67 10⁻¹¹ 1.67 10⁻²⁷ 9.1 10⁻³¹ / (0.5 10⁻³)²

           F_{g} = 4.1 10⁻⁵¹ N

           

electric force

           F_{e} = -e E

           F_{e} = - 1.6 10⁻¹⁹ 145.6

           F_{e} = 2.3 10⁻¹⁷ N

let's look for the reason between these two forces

         F_{e} / F_{g} = 2.3 10⁻¹⁷ / 4.1 10⁻⁵¹

          F_{e} /F_{g} = 10³⁴

We see that the electric force is many orders of magnitude higher than the gravitational force.

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Four identical metallic objects carry the following charges 1.82 6.65 4.80 and 9.30 C The objects are brought simultaneously int
GuDViN [60]

Answer:

a) 5.64 C

b) 3.5*10¹⁹ protons

Explanation:

a)

  • Since the four metallic objects are identical, and total charge must be conserved, this means that after brought simultaneously into contact so that each touches the others, once separated, total charge must be the same than before being brought in contact.
  • But due they are identical, after charges were able to transfer freely between them, the four objects must have the same final charge, i.e. the fourth part of the total charge, as follows:

       Q_{n} = \frac{Q_{tot}}{4} = \frac{22.57C}{4} = 5.64 C  (1)

b)

  • This charge will be divided between n protons, since the charge is positive.
  • Since each proton carries a charge equal to the elementary charge e, which value is 1.6*10⁻¹⁹ C, we can find the number of protons in excess, doing the following calculation:
  • n_{p} =\frac{Q_{n}}{e} = \frac{5.64C}{1.6e-19C} = 3.5 e19 C (2)

7 0
3 years ago
Gravity is dependent on which of the two factors?
k0ka [10]
Gravity is dependent on Mass & Distance

7 0
3 years ago
a pennyfarthing is a style of a bicycle with a very large front wheel and a small real wheel, the cyclist who sit high above and
s344n2d4d5 [400]

Answer:

In one rotation, the large wheel turns 4m.

Explanation:

The given values are:

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Mechanical advantage,

= 0.16

As we know,

⇒ Out. \ Distance = \frac{Inp. \ distance}{Mechanical \ advantage}

On putting the values, we get

⇒                         =\frac{0.64}{0.16}

⇒                         =4 \ m

4 0
3 years ago
A mass M tied to a light string is moving in a vertical circle. The tension in the string at the top is TT and TB at the bottom
natita [175]

Answer:

Explanation:

Tension provides centripetal force in the circular motion . In circular motion work done by force = torque x angle

torque is zero as , centripetal force passes through axis of rotation that is center.

So work done by centripetal force  = 0

So work done by tension on M = 0

5 0
2 years ago
Help what is the answer
Ainat [17]

Answer:

C

Explanation:

F=k\dfrac{Q_1Q_2}{r^2}= \\\\(8.99 \times 10^9)\dfrac{(-2\times 10^{-4})(8\times 10^{-4})}{0.3^2}\approx 1.6\times 10^4 N

Hope this helps!

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2 years ago
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