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Step-by-Step Solution
Step 1: Identify the Essential Concept
When two conductors are connected by a conducting wire, they attain the same electric potential. For spherical conductors, the potential on a sphere of radius $R$ carrying charge $Q$ is given by:
$V = \frac{kQ}{R},$
where $k$ is a constant ($k = \frac{1}{4\pi \varepsilon_0}$ in SI units).
Step 2: Equate the Potentials of Both Spheres
Let the radii of the two spheres be $R_A = 1 \,\mathrm{mm}$ and $R_B = 2 \,\mathrm{mm}$. If $Q_A$ and $Q_B$ are the charges on spheres $A$ and $B$ respectively at equilibrium, then:
$\frac{k Q_A}{R_A} = \frac{k Q_B}{R_B}.
$
Since $k$ is common, it cancels out, giving:
$\frac{Q_A}{R_A} = \frac{Q_B}{R_B} \quad \Rightarrow \quad \frac{Q_A}{Q_B} = \frac{R_A}{R_B} = \frac{1}{2}.
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Step 3: Express the Electric Fields on the Surface of Each Sphere
The electric field on the surface of a charged sphere of radius $R$ with charge $Q$ is:
$E = \frac{kQ}{R^2}.
$
Hence, for sphere $A$ (radius $R_A$, charge $Q_A$):
$E_A = \frac{k\,Q_A}{R_A^2}.
$
For sphere $B$ (radius $R_B$, charge $Q_B$):
$E_B = \frac{k\,Q_B}{R_B^2}.
$
Step 4: Find the Ratio of the Electric Fields
$\frac{E_A}{E_B} = \frac{\frac{k\,Q_A}{R_A^2}}{\frac{k\,Q_B}{R_B^2}}
= \frac{Q_A}{Q_B} \times \frac{R_B^2}{R_A^2}.
$
Substitute $\frac{Q_A}{Q_B} = \frac{1}{2}$ and $R_A = 1$ mm, $R_B = 2$ mm:
$\frac{E_A}{E_B}
= \frac{1}{2} \times \frac{(2 \,\mathrm{mm})^2}{(1 \,\mathrm{mm})^2}
= \frac{1}{2} \times \frac{4}{1}
= 2.
$
Therefore,
$E_A : E_B = 2 : 1.
$
Step 5: Final Answer
The ratio of the magnitudes of the electric fields at the surfaces of spheres $A$ and $B$ is $2:1$, which corresponds to the correct option.
