Final answer:
The ratio of the magnitudes of electric fields at the surfaces of two connected spheres in equilibrium is 4:1, which corresponds to their radii squared, due to the electric field being inversely proportional to the square of the radius. The correct answer is A. 4 : 1.
Step-by-step explanation:
When two spherical conductors A and B with radii 1 mm and 2 mm respectively, are connected by a conducting wire, they form an equipotential system, meaning they must have the same electric potential. Since they are conductors, when connected, they redistribute their charges until each has the same potential.
Given that the potential V for a sphere is given by the formula V = kQ/R (where k is Coulomb's constant, Q is the charge, and R is the radius), and both spheres are at the same potential after being connected, the relation between their charges will be QA/RA = QB/RB.
For spheres with radii RA = 1 mm and RB = 2 mm, the ratio of their charges in equilibrium would be QA/QB = RA/RB = 1/2.
The electric field E at the surface of a sphere is given by E = kQ/R2, so EA/EB would be (RB/RA)2 = (2 mm / 1 mm)2 = 4. Therefore, the ratio of the magnitudes of the electric fields at the surfaces of spheres A and B is 4 : 1, corresponding to option A.