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In-s [12.5K]
3 years ago
7

How are base units and derived units related?

Physics
1 answer:
Vlad1618 [11]3 years ago
8 0

Answer:

SI derived units

Other quantities, called derived quantities, are defined in terms of the seven base quantities via a system of quantity equations. The SI derived units for these derived quantities are obtained from these equations and the seven SI base units.

Explanation:

Hope this helps :D

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A hollow metal sphere has 7 cm and 11 cm inner and outer radii, respectively. The surface charge density on the inside surface i
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Answer:

1)  E = 0 , 2) zero, 3)    E = - 2,162 10⁴ N/C , 4) directed towards the center of the sphere , 5) E = 1.412 10⁴ N / C , 6)direction coming out of the sphere

Explanation:

The electric field is a vector quantity, therefore we can calculate the field due to each charge distribution and add vector  

          E = E₁ + E₂

To calculate each field we can use Gauss's law, which states that the flow is equal to the charge  by the Gaussian surface divided by ε₀

For this case, let's take a sphere as a Gaussian surface

          Ф = ∫E dA = q_{int} /ε₀

 The area of ​​a sphere is

        A = 4π r²

         E = 1 / 4πε₀   q_{int} / r²

1) r = 4 cm

This radius is smaller than the radius of the sphere, therefore the charge inside is zero and therefore the field is zero

            E = 0

2) there is no field

3) r = 8 cm

Let's calculate each field, for the inner surface

This radius is larger than the internal radius, so the field is

           σ = q_{int} / A

The area of ​​the sphere is

          V = 4 π R_in²

         Rho = q_{int} / 4π R_in²

          q_{int} = ρ 4π R_in²

         E₁ = 1 /ε₀ ρ r_in² / r²

For the outer surface

This radius is smaller so there is no load inside the Gaussian surface and therefore the field is zero

          E₂ = 0

Total E

         E = E₁ + 0

          E = 1 /ε₀  ρ₁ R_in² / r²

Let's calculate

           E = 1 /8,854 10⁻¹²   250 10⁻⁹ (7/8)²

           E = - 2,162 10⁴ N / C

4) as the electric field is negative, it is directed towards the center of the sphere

5) r = 12 cm

In this case the two surfaces contribute to the electric field,

Inner surface

        Q₁ = ρ₁ 4π R_in²

        E₁ = 1 /ε₀  ₁rho1 R_in² / r²

Outer surface

         Q₂ = ρ₂ 4π R_out²

          E₂ = 1 /ε₀  ρ₂ R_out² / r²

The total field is

          E = E₁ + E₂

          E = 1 /ε₀  | ρ | [- R_in² + R_out²] / r²

Let's calculate

          E = 1 /8,854 10⁻¹² 250 10⁻⁹ [- 7² + 11²] / 12²

          E = 1.412 10⁴ N / C

6) As the field is positive, it is directed radially with direction coming out of the sphere

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