Variation of Electric field intensity due to a uniformly charged cond…

The correct answer is Option (2)

Apply Gauss's law to a spherical Gaussian surface of radius \( r \) concentric with the shell; let the shell carry charge \( q \).

For \( r \lt R \) the Gaussian surface encloses no charge, because the whole charge sits on the shell:

    \( \displaystyle E \cdot 4\pi r^{2} = 0 \;\Rightarrow\; E = 0 \)

For \( r \ge R \) it encloses the whole charge \( q \):

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

So the field is zero inside, jumps to its largest value \( q/(4\pi\varepsilon_0 R^{2}) \) at the surface and then falls as \( 1/r^{2} \) outside. Graph (B) shows exactly this.

Why the other options are incorrect

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  Option (1) — Incorrect. The field rising linearly inside is the graph for a uniformly charged solid (non-conducting) sphere; inside a shell the field is zero.

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  Option (3) — Incorrect. The field is not zero inside, and outside it falls as \( 1/r^{2} \), not along a straight line to zero.

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  Option (4) — Incorrect. The field is not zero inside, and outside it falls as \( 1/r^{2} \), not linearly.

Note: A similar question was also asked in CUET (UG) 2023 on 12-06-2023 (Slot 2) and on 10-06-2023 (Slot 1), and in CUET (UG) 2022 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. … the field due to a uniformly charged thin shell is zero at all points inside the shell.”