What is the electric field at a point inside the shell, at a distance…

The correct answer is Option (1) → zero

Take a spherical Gaussian surface of radius \( r \lt R \), with its centre at the centre of the shell. All the charge of the shell lies outside this surface, so the enclosed charge is zero. The field has the same size at every point of this surface, so Gauss’s law gives

    \( E\times4\pi r^2=0 \)

    \( E=0 \)

Option (1) zero — Correct. The Gaussian surface inside the shell encloses no charge. So the field at every point inside the shell is zero.

Option (2) \( \frac{q}{4\pi\varepsilon_0 (r^2+R^2)} \) — Incorrect. No arrangement of the shell gives a field of this form. Inside the shell the field is zero for every \( r \).

Option (3) \( \frac{q}{4\pi\varepsilon_0 R^2} \) — Incorrect. This is the field just outside the surface of the shell, at \( r=R \). It does not hold at points inside.

Option (4) \( \frac{q}{4\pi\varepsilon_0 r^2} \) — Incorrect. This is the field at a point outside the shell, \( r \gt R \). There the whole charge \( q \) is enclosed and acts as if placed at the centre.

NCERT (2026-27):
Field due to a uniformly charged thin spherical shell – “The flux through the Gaussian surface, calculated as before, is E × 4 π r². However, in this case, the Gaussian surface encloses no charge. … that is, the field due to a uniformly charged thin shell is zero at all points inside the shell.”