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Q. Statement I: The surface charge densities of two spherical conductors of different radii are equal. Then the electric field intensities near their surface are also equal.
Statement II:Surface charge density is equal to charge per unit area.

Electric Charges and Fields

Solution:

As $\sigma_{1}=\sigma_{2}$ (Given)
$\therefore \frac{q_{1}}{4 \pi r_{1}^{2}}=\frac{q_{2}}{4 \pi r_{2}^{2}}$, or $\frac{q_{1}}{q_{2}}=\frac{r_{1}^{2}}{r_{2}^{2}}$ [Let $r_{1}$ and $r_{2}$ be two different radii]
Then the ratio of electric field intensities near the surface of spherical conductor,
$\frac{E_{1}}{E_{2}}=\frac{q_{1}}{4 \pi \varepsilon_{0} r_{1}^{2}} \times \frac{4 \pi \varepsilon_{0} r_{2}^{2}}{q_{2}}=\frac{q_{1}}{q_{2}} \times \frac{r_{2}^{2}}{r_{1}^{2}}=1$, i.e., $E_{1}=E_{2}$