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Basile [38]
4 years ago
6

A thick conducting spherical shell, with inner radius R1 and outer radius R2, is in electrostatic equilibrium. The area inside a

nd outside the spherical shell is vacuum. The net charge on the spherical shell is Q. Note: for parts (a)-(c), describe the direction of the electric field assuming that Q is positive.
(a) Find the electric field outside the shell, at a distance d > R2 from the center of the shell.
(b) Find the electric field in the shell material, at a distance d from the center of the shell where R1 < d < R2.
(c) Find the electric field at a distance d < R1 from the center of the shell, i.e. in the empty space inside the shell.
(d) How much charge is on the inner surface of the conducting shell
Physics
1 answer:
san4es73 [151]4 years ago
4 0

Answer:

a)     E = k Q / r² ,      k = 1 / 4πε₀,  b)  E=0, C) E=0,   d) Q=0

Explanation:

To find the electric field in this exercise let's use Gauss's law

         Ф = ∫ E. dA = q_{int} /ε₀

Let's use a sphere as a Gaussian surface, in this case the field lines and the radii of the spheres are parallel so the scalar product is reduced to the algebraic product, let's carry out the integral

          E A = q_{int} /ε₀

           

The area of ​​a sphere is

         A = 4π r²

we substitute

          E (4π r²) =q_{int} /ε₀

a) The field is requested outside the spherical shell

        d> R₂

in this case the charge inside the Gaussian surface is

          q_{int} = Q

           E = Q / (4π r²) ε₀

           E = k Q / r²

           k = 1 / 4πε₀

b) In the second case, the field inside the spherical casing is requested.

As the surface is metallic, the charge is located on the surface of it, the electrons repel each other. So the charge inside is zero

           q_{int} = 0

            E = 0

c) The field in the interior region of the shell, all the charge is external, therefore the internal charge is zero

           q_{int} = 0

           E = 0

d) as the shell is conductive, the electrons can move freely, which is why it moves as far away as possible between them, this means that everything is located as close as possible to the outer surface of the conductor. Consequently, THERE ARE NO ELECTRONS ON THE INTERNAL SURFACE

              Q_internal =0

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A 3 kg bowling ball is thrown onto a mattress. If it takes 0.3 seconds to stop the ball using a force of 24 N, what was the init
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Answer:

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A block of mass M=10 kg is on a frictionless surface as shown in the photo attached. And it's attached to a wall by two springs
nordsb [41]

a.

  • i. the speed of the block of mass when the springs are connected in parallel is 7.07 A m/s
  • ii. the angular velocity when the two springs are in parallel is 7.07 rad/s

b.

  • i. the speed of the block of mass when the springs are connected in series is 11.2 A m/s
  • ii. the angular velocity when the two springs are in series is 11.2 rad/s

<h3>a. </h3><h3>i. How to calculate the velocity of the mass when the springs are connected in parallel?</h3>

Since k is the spring constant of both springs = 250 N/m. The equivalent spring constant in parallel is k' = k + k

= 2k

= 2 × 250 N/m

= 500 N/m

Now since A is the maximum distance the block is pulled from its equilibrium position, the total energy of the block is E = 1/2kA

Also, 1/2k'A² = 1/2k'x² + 1/2Mv² where

  • k' = equivalent spring constant in parallel = 500 N/m,
  • A = maximum displacement of spring,
  • x = equilibrium position = 0 m,
  • M = mass of block = 10 kg and
  • v = speed of block at equilibrium position

Making v subject of the formula, we have

v = √[k'(A² - x²)/M]

Substituting the values of the variables into the equation, we have

v = √[k'(A² - x²)/M]

v = √[500 N/m(A² - (0)²)/10]

v = √[50 N/m(A² - 0)]

v = [√50]A m/s

v = [5√2] A m/s

v = 7.07 A m/s

So, the speed of the block of mass when the springs are connected in parallel is 7.07 A m/s

<h3>ii. The angular velocity of mass when the springs are in parallel</h3>

Since velocity of spring v = ω√(A² - x²) where

  • ω = angular velocity of spring,
  • A = maximum displacement of spring and
  • x = equilbrium position of spring = 0 m

Making ω subject of the formula, we have

ω = v/√(A² - x²)

Since v = 7.07 A m/s

Substituting the values of the other variables into the equation, we have

ω = v/√(A² - x²)

ω = 7.07 A m/s/√(A² - 0²)

ω = 7.07 A m/s/√(A² - 0)

ω = 7.07 A m/s/√A²

ω = 7.07 A m/s/A m

ω = 7.07 rad/s

So, the angular velocity when the two springs are in parallel is 7.07 rad/s

<h3>b. </h3><h3>i. How to calculate the velocity of the mass when the springs are connected in series?</h3>

Since k is the spring constant of both springs = 250 N/m. The equivalent spring constant in parallel is 1/k" = 1/k + 1/k

= 2/k

⇒ k" = k/2

k" = 250 N/m ÷ 2

= 125 N/m

Now since A is the maximum distance the block is pulled from its equilibrium position, the total energy of the block is E = 1/2kA

Also, 1/2k"A² = 1/2k"x² + 1/2Mv'² where

  • k" = equivalent spring constant in series = 125 N/m,
  • A = maximum displacement of spring,
  • x = equilibrium position = 0 m,
  • M = mass of block = 10 kg and
  • v' = speed of block at equilibrium position

Making v subject of the formula, we have

v = √[k"(A² - x²)/M]

Substituting the values of the variables into the equation, we have

v = √[k"(A² - x²)/M]

v = √[125 N/m(A² - (0)²)/10]

v = √[125 N/m(A² - 0)]

v = [√125]A m/s

v = [5√5] A m/s

v = 11.2 A m/s

So, the speed of the block of mass when the springs are connected in series is 11.2 A m/s

<h3>ii. The angular velocity of the mass when the springs are in series</h3>

Since velocity of spring v = ω√(A² - x²) where

  • ω = angular velocity of spring,
  • A = maximum displacement of spring and
  • x = equilbrium position of spring = 0 m

Making ω subject of the formula, we have

ω = v/√(A² - x²)

Since v = 11.2 A m/s

Substituting the values of the other variables into the equation, we have

ω = v/√(A² - x²)

ω = 11.2 A m/s/√(A² - 0²)

ω = 11.2 A m/s/√(A² - 0)

ω = 11.2 A m/s/√A²

ω = 11.2 A m/s/A m

ω = 11.2 rad/s

So, the angular velocity when the two springs are in series is 11.2 rad/s

Learn more about speed of block of mass here:

brainly.com/question/21521118

#SPJ1

3 0
2 years ago
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