The electric field just OUTSIDE a uniformly charged conducting sphere of total charge $Q$ and radius $R$, at a point distance $r > R$ from the centre, is

1. Apply Gauss's law with a spherical Gaussian surface of radius $r > R$, concentric with the conductor. 2. By spherical symmetry, $\vec{E}$ is radial and has the same magnitude on the Gaussian surface. 3. $E\,(4\pi r^2) = Q_\text{enc}/\varepsilon_0 = Q/\varepsilon_0$. 4. Solve: $E = \dfrac{Q}{4\pi\varepsilon_0\,r^2} = \dfrac{kQ}{r^2}$. 5. So the field outside is identical to that of a POINT charge $Q$ at the centre — a powerful symmetry result (newton-style 'shell theorem' for electrostatics). 6. Option B uses $R$ instead of $r$. Option C is wrong outside the conductor — shielding holds inside, not outside. Option D has wrong distance dependence. _Source: NCERT Class 12 Physics Part 1, Ch 1, §1.14 (Field due to charged sphere — Gauss's law application), p. 32._