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Katen [24]
2 years ago
13

An electron is a negatively charged particle that has a charge of magnitude, e - 1.60 x 10-19 C. Which one of the following stat

ements best describes the electric field at a distance r from the electron? The electric field is directed toward the electron and has a magnitude of ke/r. The electric field is directed away from the electron and has a magnitude of ke/2. The electric field is directed toward the electron and has a magnitude of ke/? The electric field is directed toward the electron and has a magnitude of ke?/r. The electric field is directed away from the electron and has a magnitude of ke/r.
Physics
1 answer:
irina [24]2 years ago
4 0

Explanation:

The charge on the electron is, q=-1.6\times 10^{-19}\ C

The electric field at a distance r from the electron is :

E=k\dfrac{q}{r^2}

Where

k is the electrostatic constant, k=\dfrac{1}{4\pi \epsilon_o}

We know that the electric field lines starts from positive charge and ends at the negative charge. Also, for a positive charge the field lines are outwards while for a negative charge the field lines are inwards.

So, the correct option is " the  electric field is directed toward the electron and has a magnitude of k\dfrac{q}{r^2}. Hence, this is the required solution.

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2 years ago
A solid uniformly charged insulating sphere has uniform volume charge density p and radius R. Apply Gauss's law to determine an
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Answer:

electric field E = (1 /3 e₀) ρ r

Explanation:

For the application of the law of Gauss we must build a surface with a simple symmetry, in this case we build a spherical surface within the charged sphere and analyze the amount of charge by this surface.

The charge within our surface is

 

     ρ = Q / V

     Q ’= ρ V '

The volume of the sphere is V = 4/3 π r³

     Q ’= ρ 4/3 π r³

The symmetry of the sphere gives us which field is perpendicular to the surface, so the integral is reduced to the value of the electric field by the area

      I E da = Q ’/ ε₀

      E A = E 4 πi r² = Q ’/ ε₀

      E = (1/4 π ε₀) Q ’/ r²

Now you relate the fraction of load Q ’with the total load, for this we use that the density is constant

     

      R = Q ’/ V’ = Q / V

How you want the solution depending on the density (ρ) and the inner radius  (r)

      Q ’= R V’

      Q ’= ρ 4/3 π r³

      E = (1 /4π ε₀) (1 /r²) ρ 4/3 π r³

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4 0
3 years ago
A 3.0-kg object moves to the right with a speed of 2.0 m/s. It collides in a perfectly elastic collision with a 6.0-kg object mo
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Answer:

The kinetic energy of the system after the collision is 9 J.

Explanation:

It is given that,

Mass of object 1, m₁ = 3 kg

Speed of object 1, v₁ = 2 m/s

Mass of object 2, m₂ = 6 kg

Speed of object 2, v₂ = -1 m/s (it is moving in left)

Since, the collision is elastic. The kinetic energy of the system before the collision is equal to the kinetic energy of the system after the collision. Let it is E. So,

E=\dfrac{1}{2}m_1v_1^2+\dfrac{1}{2}m_2v_1^2

E=\dfrac{1}{2}\times 3\ kg\times (2\ m/s)^2+\dfrac{1}{2}\times 6\ kg\times (-1\ m/s)^2

E = 9 J

So, the kinetic energy of the system after the collision is 9 J. Hence, this is the required solution.

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How do fluids excert pressure​
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All fluids exert pressure like the air inside a tire. The particles of fluids are constantly moving in all directions at random. As the particles move, they keep bumping into each other and into anything else in their path. These collisions cause pressure, and the pressure is exerted equally in all directions.

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What additional information do you need to prove ∆ABC ≅ ∆DEF by the SAS Postulate?
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Answer:

Option A

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