The mass of Jupiter is 1.9 × 1027 and that of the sun is 1.99 × 1030. The
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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 = /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