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lana66690 [7]
3 years ago
14

A +5.00 pC charge is located on a sheet of paper.

Physics
1 answer:
emmainna [20.7K]3 years ago
8 0

Answer:

a)    V = - x ( σ / 2ε₀)

c)  parallel to the flat sheet of paper

Explanation:

a) For this exercise we use the relationship between the electric field and the electric potential

          V = - ∫ E . dx        (1)

for which we need the electric field of the sheet of paper, for this we use Gauss's law. Let us use as a Gaussian surface a cylinder with faces parallel to the sheet

       Ф = ∫ E . dA = q_{int} /ε₀

the electric field lines are perpendicular to the sheet, therefore they are parallel to the normal of the area, which reduces the scalar product to the algebraic product

          E A = q_{int} /ε₀

area let's use the concept of density

        σ = q_{int}/ A

       q_{int} = σ A

          E = σ /ε₀

as the leaf emits bonnet towards both sides, for only one side the field must be

          E = σ / 2ε₀

         we substitute in equation 1 and integrate

      V = - σ x / 2ε₀  

       V = - x ( σ / 2ε₀)

if the area of ​​the sheeta is 100 cm² = 10⁻² m²

      V = - x  (10⁻²/(2 8.85 10⁻¹²) = - x  ( 5.6 10⁻¹⁰)

       

      x = 1 cm     V = -1   V

      x = 2cm     V = -2   V

This value is relative to the loaded sheet if we combine our reference system the values ​​are inverted

       V ’= V (inf) - V

       x = 1 V = 5

       x = 2 V = 4

       x = 3 V = 3

   

These surfaces are perpendicular to the electric field lines, so they are parallel to the sheet.

 

In the attachment we can see a schematic representation of the equipotential surfaces

b) From the equation we can see that the equipotential surfaces are parallel to the sheet and equally spaced

c) parallel to the flat sheet of paper

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Answer:

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Hi there!

a) Let´s calculate how much distance it is the last 11% of the fall:

96 m · 0.11 = 10.56 m

So, we have to find how much time it takes the bolt to pass from a height of 10.56 m to the ground.

First, let´s calculate how much time it takes the bolt to reach a height of 10.56 m. For that we can use this equation:

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g = acceleration due to gravity.

If we consider the ground as the origin of the frame of reference, then h0 = 96 m. Since the bolt is dropped, the initial velocity is zero (v0 = 0). Then, the equation gets reduce to this:

h = h0 + 1/2 · g · t²

We have to find at which time h = 10.56 m.

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Now that we have the time at which the bolt is located at 10.56 m above the ground, we can calculate the velocity of the bolt at that time.

The equation of velocity (v) of the bolt is the following:

v = v0 + g · t

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v = 0 - 9.8 m/s² · 4.2 s

v = -41.2 m/s

<u>When the bolt begins to fall the last 11% of the fall its velocity is -41.2 m/s.</u>

Now, we can calculate how much time it takes to fall the last 10.56 m.

The initial velocity of the bolt will be the velocity at h = 10.56 m. The initial height will be 10.56 m.

h = h0 + v0 · t + 1/2 · g · t²

We have to find the time at which h = 0 (the bolt hits the ground)

0 = 10.56 m - 41.2 m/s · t - 1/2 · 9.8 m/s² · t²

Solving the quadratic equation using the quadratic formula:

t = 0.25 s (the other solution of the quadratic equation is negative and thus discarded).

<u>It takes the bolt 0.25 s to pass the last 11% of the fall.</u>

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v = v0 + g · t

v = -41.2 m/s - 9.8 m/s² · 0.25 s

v = -43.6 m/s

<u>The velocity of the bolt just before it reaches the bolt is -43.6 m/s</u>

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