1. Key principle when conductors are connected

   When two conducting spheres are connected by a conducting wire, charge flows until their electric potentials become equal.

   So, at equilibrium: $V_1 = V_2$

2. Potential of a charged conducting sphere

   The potential of an isolated conducting sphere of radius $R$ carrying charge $Q$ is: $V = \frac{1}{4\pi\varepsilon_0}\frac{Q}{R}$

   For the two spheres:

   • Sphere 1: radius $a$, charge $q_1$
   • Sphere 2: radius $b$, charge $q_2$

   Their potentials are: $V_1 = \frac{1}{4\pi\varepsilon_0}\frac{q_1}{a}$ $V_2 = \frac{1}{4\pi\varepsilon_0}\frac{q_2}{b}$

3. Set potentials equal

   Since they are connected: $\frac{1}{4\pi\varepsilon_0}\frac{q_1}{a} = \frac{1}{4\pi\varepsilon_0}\frac{q_2}{b}$

   Cancelling the common factor $\frac{1}{4\pi\varepsilon_0}$: $\frac{q_1}{a} = \frac{q_2}{b}$

4. Find the ratio of charges

   Rearranging: $\frac{q_1}{q_2} = \frac{a}{b}$

   Therefore, the ratio of charges on the two spheres respectively is: $q_1 : q_2 = a : b$

5. Check options

   • A: $ab$ ❌
   • B: $\frac{b}{a}$ ❌
   • C: $\frac{a}{b}$ ✅
   • D: $\sqrt{ab}$ ❌

Final Answer: Option C