Gauss's Law

Gauss's Law for a spherical Gaussian surface of radius (r) with (R < r < 2R) states (\oint \vec{E} \cdot d\vec{A} = Q_{enc}/\varepsilon_0). The key is determining the enclosed charge. The solid conducting sphere of radius (R) has charge (+Q), all residing on its outer surface at (r = R). The thin shell at (r = 2R) has charge (-Q) on it, but it lies outside the Gaussian surface. Therefore, for (R < r < 2R), the enclosed charge is (Q_{enc} = +Q). By spherical symmetry, (E \cdot 4\pi r^2 = Q/\varepsilon_0), giving (E = kQ/r^2) directed radially outward. Option directed radially inward with (kQ/4r^2) is incorrect because it wrongly suggests the outer shell affects this region. Option 'zero' is incorrect because the outer shell's charge is not enclosed and cannot cancel the inner sphere's contribution. Option (2kQ/r^2) inward has no physical basis. This is a multi-conductor Gauss's Law problem standard in JEE Advanced. Plausibility check: by shell theorem, the outer shell at (2R) exerts no field inside it, so only the inner sphere contributes, confirming (E = kQ/r^2).