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An electron is traveling with initial kinetic energy K in a uniform electric field. The electron comes to rest momentarily after traveling a distance d.

a. What is the magnitude of the electric field?
b. What is the direction of the electric field?

1. In the direction of the electron's motion.
2. Opposite to the direction of the electron's motion.
3. Perpendicular to the direction of the electron's motion

c. What If? Fluoride ions (which have the same charge as an electron) are initially moving with the same speed as the electrons from part (a) through a different uniform electric field. The ions come to a stop in the same distance d. Let the mass of an ion be M and the mass of an electron be m. Find the ratio of the magnitude of electric field the ions travel through to the magnitude of the electric field found in part (a). (Use the following as necessary: d, K, m, M, and e for the charge of the electron.)

Respuesta :

Answer:

a. The magnitude of the electric field is [tex]E = -\dfrac{K}{e \cdot S}[/tex]

b. 1. In the direction of the electron's motion

c. The ratio of the magnitude of the electric field the ions travel through to the magnitude of electric field for the electrons is M:m or 34634.6:1

Explanation:

The change in kinetic energy of electron = [tex]\dfrac{1}{2} \cdot m \cdot v^2_f - \dfrac{1}{2} \cdot m \cdot v^2_i[/tex]

Given that the work done, W = the change in kinetic energy of electron, we have;

[tex]W = \dfrac{1}{2} \cdot m \cdot v^2_f - \dfrac{1}{2} \cdot m \cdot v^2_i[/tex]

Given that the final velocity is 0, we have;

[tex]W =- \dfrac{1}{2} \cdot m \cdot v^2_i = -K[/tex]

However, we have;

W = E·q·S

Where:

q = Particle charge = Electron charge = e

S = Particle displacement

E = Electric field strength

Therefore;

W = -K = E·q·S

Which gives;

[tex]E = -\dfrac{K}{e \cdot S}[/tex]

b. The direction of the electric field is the direction of the force exerted by the electric field on a positive charge

Given that the particle is an electron, its direction is opposite to that of a positive charge, however whereby the electron is brought to rest the direction of the field is in the same direction of as that of the electron's motion

The correct option is therefore, 1. In the direction of the electron's motion

c. Given that the mass of fluoride ion, Fe⁻, M = 3.155×10⁻²⁶ kg

The mass of an electron, m = 9.1094×10⁻³¹ kg

We have;

x × 0.5 × m × v² = 0.5×M×v²

∴ x = M/m

Therefore. the kinetic energy of the fluoride ion is M/m multiplied by the kinetic energy of the electron

x × 0.5 × 9.1094×10⁻³¹ × v² = 0.5×3.155×10⁻²⁶×v²

Therefore, x = (3.155×10⁻²⁶)÷(9.1094×10⁻³¹) = 34634.6

Therefore, the kinetic energy of the Fluoride ion is 34634.6 times that of the kinetic energy of the electron

Whereby the electric field strength is directly proportional to the kinetic energy, we have;

[tex]E_{(electron)} = -\dfrac{K_{(electron)}}{e \cdot S}[/tex]

[tex]E_{(fluoride \ ion)} = -\dfrac{K_{(fluoride \ ion)}}{e \cdot S}[/tex]

Which gives;

[tex]\dfrac{E_{(fluoride \ ion)} }{E_{(electron)}} = \dfrac{-\dfrac{K_{(fluoride \ ion)}}{e \cdot S}}{ -\dfrac{K_{(electron)}}{e \cdot S}} = \dfrac{{K_{(fluoride \ ion)}}}{ {K_{(electron)}}}[/tex]

[tex]\dfrac{E_{(fluoride \ ion)} }{E_{(electron)}} = \dfrac{\dfrac{M}{m} \times {K_{(electron)}}}{ {K_{(electron)}}} = \dfrac{M}{m}[/tex]

[tex]\dfrac{E_{(fluoride \ ion)} }{E_{(electron)}} = \dfrac{34634.6 \times {K_{(electron)}}}{ {K_{(electron)}}} = 34634.6[/tex]

The magnitude of the uniform electric field in the fluoride ion path is 34634.6 times that of the magnitude of the electric field in the electron path.

The ratio of the magnitude of the electric field the ions travel through to the magnitude of electric field for the electrons is M:m or 34634.6:1

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