JEE Main 2024 — Electrostatics Question with Solution

When two conducting spheres of radii $$a$$ and $$b$$ are connected by a conducting wire, they come to the same potential because conductors in contact share charges until their potentials become equal. The potential of a charged sphere is given by $$V=\frac{kQ}{R}$$, where $$V$$ is the potential, $$k$$ is Coulomb's constant, $$Q$$ is the charge on the sphere, and $$R$$ is the radius of the sphere.

For the two spheres at the same potential, we have:

$$\frac{k{Q}_a}{a}=\frac{k{Q}_b}{b}$$

Where:

• $$Q_a$$ and $$Q_b$$ are the charges on the spheres with radii $$a$$ and $$b$$, respectively.
• $$k$$ cancels out from both sides as it is a constant.

From the above equation, to find the ratio of charges $$\frac{Q_a}{Q_b}$$, we rearrange it as follows:

$$\frac{Q_a}{Q_b}=\frac{a}{b}$$

Therefore, the ratio of the charges of the two spheres respectively is $$\frac{a}{b}$$.

So, the correct answer is Option C: $$\frac{a}{b}$$.