The electric field due to a uniformly charged spherical shell of radi…

The correct answer is Option (3)

Apply Gauss’s law with a spherical Gaussian surface of radius r concentric with the shell.

Inside (\( r\lt R \)): the surface encloses no charge, so \( E = 0 \).

Outside (\( r\ge R \)): the surface encloses the whole charge q, so the field is that of a point charge at the centre:

    \( \displaystyle E = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^{2}} \)

The graph is therefore zero up to \( r = R \), jumps to the value \( q/4\pi\varepsilon_0R^{2} \) at the surface, and then falls as \( 1/r^{2} \). This is the graph in Option (3).

Why the other options are incorrect

• 

  Option (1) — Incorrect. It shows a constant non-zero field inside the shell, but Gauss’s law gives zero there.

• 

  Option (2) — Incorrect. The field rising linearly inside is that of a uniformly charged solid (non-conducting) sphere, where charge is enclosed at every radius.

• 

  Option (4) — Incorrect. It shows a field that stays constant outside the shell and is non-zero inside the shell, but the field outside must fall as \( 1/r^{2} \).

Note: A similar question was also asked in CUET (UG) 2023 on 10-06-2023 (Slot 1), and in CUET (UG) 2022 on 23-08-2022 (Slot 1) and on 18-08-2022 (Slot 2).

NCERT (2026-27):
Field Due to a Uniformly Charged Thin Spherical Shell – “… 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.”

Field Due to a Uniformly Charged Thin Spherical Shell – “… However, in this case, the Gaussian surface encloses no charge. Gauss’s law then gives

    \( \displaystyle E \times 4\pi r^2 = 0 \quad \text{i.e.,} \quad E = 0 \quad (r \lt R) \qquad (1.35) \)

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