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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 [tex]\phi_E[/tex] out of the surface and the magnitude E of the electric field at the surface as a result of these doublings?a. [tex]\phi_E[/tex] and E do not change.b. [tex]\phi_E[/tex] increases and E remains the same.c. [tex]\phi_E[/tex] increases and E decreases.d. [tex]\phi_E[/tex] increases and E increases.

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IamJerry

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.      

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