JEE Main 2025 — Electrostatics Question with Solution

When two conducting spheres are in contact, they achieve the same electric potential. Consider two metal spheres with radii $R$ and $3R$ that have the same surface charge density $\sigma$.

For a conductive sphere, the potential $V$ of a sphere can be expressed as $V=\frac{\sigma r}{{\epsilon }_0}$, where $r$ is the radius of the sphere and ${\epsilon }_0$ is the permittivity of free space.

When the two spheres are brought into contact and then separated, they equalize their potentials:

$V_1=V_2$

Thus, the equations become:

${\sigma }_1{r}_1={\sigma }_2{r}_2$

Substituting for the radii:

${\sigma }_{1}R={\sigma }_2(3R)$

From this, we can solve for the ratio of the surface charge densities:

$\frac{{\sigma }_1}{{\sigma }_2}=\frac{3R}{R}=3$

Therefore, the ratio of the surface charge densities on the smaller sphere to the larger sphere after contact is 3.