Answer:
a) q_shell = -Q , b) q_shell = -Q
Step-by-step explanation:
a) We can solve this exercise using Gauss's law
Ф = ∫ E. dA =
/ε₀
For this we use a Gaussian surface that takes advantage of the symmetry of the problem, we select a sphere that this force of the spherical plastic shell.
In this case the electric field lines coincide with the radii of the sphere and the scalar product reduces to the algebraic product, the area of the sphere
A = 4 π r²
fi = E 4π r² = q_{int} /ε₀
apply this formula to our case, we have that the charge inside the Gaussian sphere is
q_{int} = q_sphere + q_shell
the charge of the sphere is Q and the requirement is that there is no electric field outside the Guasian sphere, for this the net charge inside it must be zero
0 = Q + q_shell
q_shell = -Q
b) In Gauss's law, only the charge inside the Gaussian surface matters, not its position, therefore if the charge is not in the center the result remains the same.
q_shell = -Q