Electric field due to a uniformly charged thin spherical shell(A) is…

The correct answer is Option (2) → (B) and (D) only

Apply Gauss's law to a spherical Gaussian surface of radius r, concentric with the shell of radius R.

    \( \displaystyle E \cdot 4\pi r^{2} = \frac{q_{\mathrm{enc}}}{\varepsilon_0} \)

(A) is directed outward, if q < 0 — Incorrect. For a negative charge the field points radially inward; it is outward only for q > 0.

(B) at a point outside the shell is same as that of electric field due to point charge. — Correct. For r > R the whole charge is enclosed, so \( E = \frac{q}{4\pi\varepsilon_0 r^{2}} \), as if all the charge were at the centre.

(C) is finite and same everywhere inside the shell. — Incorrect. Inside the shell the field is zero, not a finite value.

(D) is zero everywhere inside the shell. — Correct. For r < R the Gaussian surface encloses no charge, so E = 0 at every inside point.

NCERT (2026-27):
Field Due to a Uniformly Charged Thin Spherical Shell – “… The electric field is directed outward if \( q \gt 0 \) and inward if \( q \lt 0 \). This, however, is exactly the field produced by a charge \( q \) placed at the centre O. Thus for points outside the shell, the field due to a uniformly charged shell is as if the entire charge of the shell is concentrated at its centre.

… However, in this case, the Gaussian surface encloses no charge. Gauss’s law then gives

    \( \displaystyle E \times 4\pi r^2 = 0 \)

i.e.,     \( E = 0 \quad (r \lt R) \qquad (1.35) \)

that is, the field due to a uniformly charged thin shell is zero at all points inside the shell.”