A spherical surface surrounds a point charge it its center. If the charge is doubled and if the radius of the surface is also doubled, what happens to the electric flux out of the surface and the magnitude E of the electric field at the surface as a result of these doublings?a. and E do not change. b. increases and E remains the same. c. increases and E decreases. d. increases and E increases.

c. Ф increases and E decreases

Explanation:

From Gauss's Law

                         Ф = Q / ε₀                    _____(1)

The electric flux Ф  is the surface integral of electric field E

                       Ф                   _____(2)

                       |E| = Q/ 4πε₀r²                 _____(3)

Where,

Φ is the electric flux through a closed surface S enclosing any volume V

E is the electric field

dA is a vector representing an infinitesimal element of area of the surface

Q is the total charge enclosed within V, and

ε₀ is the electric constant

Considering the relationship between equation 1, 2 and 3, If the charge, Q, is doubled and the radius, r, of the surface is also doubled; the electric flux Φ increases and electric field E decreases.  From Gauss law in equation 1, the electric flux Φ is directly proportional to the charge, Q, with ε₀ being a constant. An increase in charge will cause an increase in electric flux. Also, the electric field, will decrease because of the inverse relationship between radius and electric field.      

i guess it’s a and b

it breaks and dirupture other materials and other sencews from the bottom