Two charged metallic spheres with radii R₁ and R₂ are brought in contact and then separated. The ratio of final charges Q₁ and Q₂ on the two spheres respectively will be _______.

The correct answer is:
$\frac{Q_2}{Q_1}=\frac{R_2}{R_1}$

Step-by-Step Explanation:

When two charged metallic (conducting) spheres are brought into contact, charge flows from one sphere to the other until they reach a common electric potential. Let this common potential be V.

The potential V on the surface of a charged conducting sphere of radius R carrying a charge Q is given by the formula:
$V=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{R}$

Let $Q_1$ and $Q_2$ be the final charges on the two spheres of radii $R_1$ and $R_2$ respectively, after they are separated.

Since the spheres were in contact, their final potentials must be equal:
$V_1=V_2$

Substituting the expression for potential for each sphere, we get:
$\frac{1}{4\pi {\epsilon }_0}\frac{Q_1}{R_1}=\frac{1}{4\pi {\epsilon }_0}\frac{Q_2}{R_2}$

Canceling the constant term $\frac{1}{4\pi {\epsilon }_0}$ from both sides, we obtain:
$\frac{Q_1}{R_1}=\frac{Q_2}{R_2}$

Rearranging the terms to find the ratio of the final charges on the two spheres, we get:
$\frac{Q_2}{Q_1}=\frac{R_2}{R_1}$

This shows that the ratio of the final charges on the spheres is directly proportional to the ratio of their radii.