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Alexus [3.1K]
3 years ago
5

What is the current in a circuit with a 9.0 V battery and a resistance of 36.0 Ω?

Physics
2 answers:
ddd [48]3 years ago
5 0
Current=voltage/resistance= 9.0 V/36.0<span>Ω=1/4=0.25 A</span>
Firlakuza [10]3 years ago
4 0
I = V / R = 9v / 36ohms = 1/4 Ampere
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A spherical shell has inner radius Rin and outer radius Rout. The shell contains total charge Q, uniformly distributed. The inte
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  E = Q / 4 π ε₀  (r-R_int) / (R_put³ -R_int³)

Explanation:

For this exercise we can use Gauss's law

          Ф = ∫ E. dA = q_{int} / ε₀

Where q is the charge inside the surface.

In this case the surface must be a sphere, the electric field lines and the radius of the sphere are parallel, so the scalar product is reduced to the algebraic product

           ∫ E dA = q_{int}/ ε₀

The area of ​​a sphere is

           A = 4π r²

- The electric field for a distance r < R_int

The charge inside is zero, so the electric field

          E = 0        r <R_in t

- The field for a radio inside the shell

   Let's use the concept of density

         ρ = Q / V

         q = ρ (4/3 π r³)

         dq = ρ 4π r² dr

We substitute in the Gaussian equation

     E ∫ dA = ρ 4π r² dr / ε₀

    E 4π r² = ρ 4π/ε₀   r³ / 3

     E = ρ / 3ε₀ r

We evaluate between the lower limit r = R_int, E = 0 and the upper limit r = r, E = E

      E- 0 = ρ / 3ε₀ (r –R_int)

Density  is

      ρ = q / 4/3 π (R_out³ - R_int³)

Where R <r

     E = Q / 4 π ε₀  (r-R_int) / (R_put³ -R_int³)

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