Answer:
c. Ф[tex]_{E}[/tex] increases and E decreases
Explanation:
From Gauss's Law
Ф[tex]_{E}[/tex] = Q / ε₀ _____(1)
The electric flux Ф[tex]_{E}[/tex] is the surface integral of electric field E
Ф[tex]_{E}[/tex] [tex]= \int\limits_{s}{E} \, dA[/tex] _____(2)
|E| = Q/ 4πε₀r² _____(3)
Where,
Φ[tex]_{E}[/tex] 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[tex]_{E}[/tex] increases and electric field E decreases. From Gauss law in equation 1, the electric flux Φ[tex]_{E}[/tex] 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.
