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 Φ_E out of the surface and the magnitude E of the electric field at the surface as a result of these doublings? A)Φ_E and E do not change. B)Φ_E increases and E remains the same. C)Φ_E increases and E decreases. D)Φ_E increases and E increases.

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Quizplus · Accepted Answer

Electric flux (Φ) depends on the charge enclosed by the surface and is given by Gauss's law as Φ = q/ε₀, where q is the charge and ε₀ is the permittivity of free space. Doubling the charge doubles the electric flux, regardless of the size of the spherical surface. The electric field (E) at the surface of a sphere surrounding a point charge is given by E = kq/r², where k is Coulomb's constant, q is the charge, and r is the radius of the sphere. Doubling both the charge and the radius results in the electric field E being halved, since E is inversely proportional to the square of the radius.

A Spherical Surface Surrounds a Point Charge It Its Center