maharjansarinya
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Two plane metal plates 4 cm long are held horizontal 3 cm apart in a vacuum one being vertically one above other. The upper plate is at a p.d 300 volt and the lowered is earthed. Electrons having velocity of 10^7 m/s are injected horizontally midway between the plate and in a direction parallel to the 4cm edge. Calculate the vertical deflection of the electron beam as it emerges from the plates. (e/m=1.8 x 10^11Ckg^- 1).​

Sagot :

With Newton's second law and kinematics we find the deflection of the electron beam is y = 1.44 cm

given parameters

  • the voltage of each upper plate V = 300 V, the lower one zero volts
  • the length of the plates, l = 4 cm (1 m / 100 cm) = 0.04 m
  • the separation between the plates d = 3 cm = 0.03 m
  • the velocity of the electrons v = 10⁷ m/s parallel to the plates
  • the relationship e / m = 1.8 10 11 C / kg

to find

  • the deflection (y) of electrons

the  electric field and the electric potential are related

         V = -E d

         E = - V / d

where E is the electric field, V the electric potential and d the separation between the plates

         E = - 300 / 0.03

         E = -1 10⁴ N / C

Let's solve this exercise in parts

1st part. Let's set a reference system with the horizontal axis in direction to the length of the plates and the y axis in the vertical direction, the zero is located at the initial point of the plate and half of its separation.

Let's use Newton's second law to find the acceleration in the y-axis of the electrons

           F = m a

electric force is

           F = q E

           q E = m a

           a = q / m E

the face of electrons is negative q = -e

            a = - e / m E

            a = - 1.8 10¹¹ (- 1 10⁴)

            a = 1.8 10¹⁵ m /s²

this acceleration is directed upwards, that is, the electrons approach the positive plate.

2nd part. We use kinematics to find the time it takes for electrons to pass the plates; on the x-axis there is no acceleration, so we use the uniform motion relationships

             v = x / t

             t = x / v

             t = 0.04 / 1 10⁷

             t = 4 10⁻⁹ s

For the y-axis, let it be known that the electrons are initially in the middle between the plates, so y₀ = 0 and the velocity eg this axis is zero, we look for the deflection at the end of the plate

          y = y₀ + v_{oy) t + ½ a t²

           y = 0 + 0 + ½ a t²

           y = ½ (1.8 10¹⁵)  ( 4 10⁻⁸ )²

           y = 14.4 10⁻³ m = 1.44 10⁻² m

          y = 1.44 cm

We can use Newton's second law and kinematics to find the distance that the electron beam deviates from the middle of the plate is y = 1.44 cm, therefore the beam of electrons leaving the plates

learn more about Newton's second law and kinematics here: brainly.com/question/13016721

Answer:

The vertical deflection of the beam of an electron beam is 1.44×〖10〗^(-2)  m.

Explanation:

Velocity (v) = 〖10〗^7  m/s

p.d. (V) = 300 V

Length of metal plates (x) = 4cm = 0.04 m

Separation (d) = 3 cm = 0.03 m

Specific charge of an electron (e/m) = 1.8×〖10〗^11  C/kg

Vertical deflection (y) = ?

The vertical deflection of electron inside the electric field is

y=1/2  at^2

=1/2  eE/m  (x/v)^2

=1/2 (e/m)  V/d  (x/v)^2

=1/2×1.8×〖10〗^11×300/0.03×(0.04/〖10〗^7 )^2

⸫ y=1.44×〖10〗^(-2)  m

Hence, the vertical deflection of the beam of an electron beam is 1.44×〖10〗^(-2)  m.

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