Respuesta :
Answer:
A) Q = Po 4/3 π R³
Explanation:
a) Let's calculate the total charge, if the charge density is
p (r) = Po
The definition of density is the charge between the volume
ρ = Q / V
The volume of the sphere is
V = 4/3 π R³
We calculate the charge
Q = ρ V
Q = Po 4/3 π R³
B) For this part of the problem we will use Gauus's law that states that the electric flow through a surface is equal to the charge within the surface between the electrical permittivity (ε₀ = 8.85 10⁻¹²2 C² / N m²)
Φ = ∫ E. dA = [tex]q_{int}[/tex] / ε₀
Bold indicate vectors
Let's apply this equation to our sphere, for this we build a spherical Gaussian surface since it has the shape of the body. In this case the electric field lines exit concentrically from the sphere, which coincides with the radii of the Gaussian sphere, bone are parallel, so the scalar product behaves like an ordinary product, the equation is
Φ = ∫ E dA = [tex]q_{int}[/tex] / ε₀
Let's calculate the field for r> R
In this case, the entire sphere charge is waxed on the Gaussian surface.
[tex]q_{int}[/tex] = Q
The integral
E ∫ d A = [tex]q_{int}[/tex] / ε₀
The area of the sphere is
A = 4 pi r²
E (4 pi r²) = Q / ε₀
E = 1 / (4πε₀) Q / r²
Let's calculate for the case r <R
In this case the Gaussian surface is within the sphere, the charge that is outside it does not contribute to the field, so we will calculate the charge within the Gaussian surface, for this we will use the concept of uniform density
ρ (r) = Po = [tex]q_{int}[/tex] / [tex]V_{int}[/tex]
Where Vint is the volume within the surface
V = 4/3 π r³
Let's replace and calculate inner charge (qint)
[tex]q_{int}[/tex]= Po [tex]V_{int}[/tex]
[tex]q_{int}[/tex] = Po (4/3 π r³)
Let's solve the integral
E A = [tex]q_{int}[/tex] /ε₀
E (4 π r²) = Po (4/3 π r³) /ε₀
E = Po r / 3ε₀
It is very common to give this elective field based on the total charge of the sphere,
For this we remember that
Po = Q / V
Po = Q / (4/3 π R³)
E = Q / (4/3 π R³) r / 3 ε₀
E = (1/4 π ε₀) Q r / R³
