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AysviL [449]
3 years ago
12

What is the process of conduction in terms of particle movement

Physics
1 answer:
ivann1987 [24]3 years ago
8 0
<span>What is the process of conduction in terms of particle movement chocies

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Answer:

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Two particles, each with charge 55.3 nC, are located on the y axis at y 24.9 cm and y -24.9cm (a) Find the vector electric field
ser-zykov [4K]

Answer:

Ex = kq 2x / ∛ (x² + y²)²  and  Ex = 2008 N / C

Explanation:

a)   The electric field is a vector quantity, so we must find the field for each particle and add them vectorially, as the whole process is on the X axis,

The equation for the electric field produced by a point charge is

         E = k q / r²

With r the distance between the point charge and the positive test charge

We look for each electric field

Particle 1.  Located at y = 24.9 m, let's use Pythagoras' theorem to find the distance

          r² = x² + y²

          E1 = k q / (x² + y²)

Particle 2.   located at x = -24.9 m

          r² = x² + y²

          E2 = k q / (x² + y²)

We can see that the two fields are equal since the particles have the same charge and coordinate it and that is squared.

In the attached one we can see that the Y components of the electric fields created by each particle are always the same and it is canceled, so we only have to add the X components of the electric fields. Let's use Pythagoras' theorem to find

Let's measure the angle from axis X

     cos θ = CA / H = x / (x2 + y2) ½

     E1x = E1 cos θ

      E2x = = E1 cos θ

The resulting field

      Ey = 0

      Ex = E1x + E2x 2 E1x

      Ex = 2 k q / (x² + y²) cos θ) = 2 k q / (x² + y²) x / √(x² + x²)

      Ex = kq 2x / ∛ (x² + y²)²

b) For this part we substitute the numerical values

      Ex = 8.99 10⁹ 55.3 10⁻⁹ x / (x² + 0.249 2) ³/₂

      Ex = 497.15   x / (x² + 0.062)  ³/₂  

Point where can the value of the electric field x = 38.1 cm = 0.381 m

       Ex = 497.15 0.381 / (0.381² + 0.062)  ³/₂  

       Ex = 497.15 0.381 / (0.1452 + 0.062) 3/2 = 189.41 / 0.2072 3/2

       Ex= 189.41 /0.0943

       Ex = 2008 N / C

c)  E = 1.00 kN / C = 1000 N / C

To solve this part we must find x in the equation

       Ex = 497.15 x / (x² + 0.062)  ³/₂  

Let's use some arithmetic

       Ex / 497.15 = x / (x² + 0.062)  ³/₂  

       [Ex / 497.15] ²/₃ = [x / (x² + 0.062) 3/2] ²/₃

       ∛[Ex / 497.15]² = (∛x²) / (x² + 0.062)                 (1)

The roots of this equation are the solution to the problem,

     

For Ex = 1.00 kN / C = 1000 N / C

 

      [Ex / 497.15] 2/3 = 1000 / 497.15) 2/3 = 1,312

       1.312 = (∛x² ) / (x² + 0.062)

       1.312 (x² + 0.062) = ∛x²

       1.312 X² - ∛x² + 1.312 0.062 = 0

       1.312 X² - ∛x² + 0.0813 = 0

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