Answer:
c. [tex]\phi_1=\phi2[/tex]
Explanation:
the electric flux is given by the Gaussian's theorem:
[tex]\int \vec{E}\cdot d\vec{S}=\frac{Q}{\epsilon_0}[/tex]
where Q is the net charge inside the surface and e0 is the dielectric permittivity of vacuum. Due to in both cases the electric flux is independt of the radius of the surface and only depends of the charge Q. We conclude that both fluxes are equal.
[tex]\phi_1=\phi2[/tex]
answer: c
hope this helps!!
