Question

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

Answer 1

Answer:

c. Ф$_{E}$ increases and E decreases

Explanation:

From Gauss's Law

                         Ф$_{E}$ = Q / ε₀                    _____(1)

The electric flux Ф$_{E}$  is the surface integral of electric field E

                       Ф$_{E}$   $= \int\limits_{s}{E} \, dA$                 _____(2)

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

Where,

Φ$_{E}$ 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$_{E}$ increases and electric field E decreases.  From Gauss law in equation 1, the electric flux Φ$_{E}$ 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.