QJEEM24Phc1f2: JEE Main 2024 Physics - Class 12 JEEM24Phc1f2 Solution: Two charged conducting spheres of radii and are connected to…

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:

$\frac{kQ_a}{a} = \frac{kQ_b}{b}$

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: $\frac{a}{b}$ .