Respuesta :
Answer:
A) E = 1/4π ε₀ r/3a³ , B) E = 1/4πε₀ Q / r²
, C) E = 0
, D) E = 1/4πε₀ 3Q / r²
Explanation:
For this exercise let's use Gauss's law
Ф = ∫ E. dA = / ε₀
Since the two objects are spherical we will use a sphere as our Gaussian surface
E A = q_{int} /ε₀
E 4π r² = q_{int} /ε₀
Part A
The radius is r <a
In this case the chage inside of the Gaussian surface can be found using the concept of density
ρ = q / V
q_{int} = ρ 4/3 π r³
We replace
E 4π r² = rho 4/3 π r³ 1/ε₀
E = rho r/3ε₀
E = 1/4π ε₀ r/3a³ r <a
Part B
In this case they ask for the field between the sphere and the shell
a <r <b
In this case the charge inside is the charge of the sphere, we replace
E (4π r²) = Q /ε₀
E = 1 / 4πε₀ Q / r²
Part C
The electric filed in the region inside the spherical shell
b <r <c
An electric conductor in equilibrium all the charge is on its surface so the charge inside the conductor is zero
q_{int}= 0
E = 0
Part D
In the region outside ofthe hollow shell
c <r
The charge inside is
q_{int} = Q + 2Q = 3 Q
The electric field is
E = 1 / 4πε₀ 3Q / r²
