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(C) Let the potentials of the spheres with radii $a$ and $b$ be $V_a$ and $V_b$ respectively,and the charges on them be $Q_a$ and $Q_b$.Since the two

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$\frac{k Q}{a+b}$

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(C) Let the potentials of the spheres with radii $a$ and $b$ be $V_a$ and $V_b$ respectively,and the charges on them be $Q_a$ and $Q_b$.Since the two spheres are connected by a conducting wire,they must be at the same potential:$V_a = V_b$$\frac{k Q_a}{a} = \frac{k Q_b}{b}$$\frac{Q_a}{Q_b} = \frac{a}{b}$Using the property of ratios,$\frac{Q_a}{Q_a + Q_b} = \frac{a}{a + b}$.Since the total charge $Q_a + Q_b = Q$,we have:$Q_a = \frac{a Q}{a + b}$ and $Q_b = \frac{b Q}{a + b}$.Now,calculating the potential $V$ of each sphere:$V = V_a = V_b = \frac{k Q_a}{a} = \frac{k}{a} \left( \frac{a Q}{a + b} \right) = \frac{k Q}{a + b}$.

Two metallic spheres of radii $a$ and $b$ are placed far away from each other and are connected by a thin conducting wire. The total charge on them is $Q$. Calculate the potential of each sphere.

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