The variation of electric field with the distance from centre due to solid uniformly charged spherical conductor is correctly represented by (Given \( R \) = Radius of spherical conductor)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2)
On a conductor the charge resides on the outer surface and the electrostatic field inside it is zero. Outside, Gauss's law with a spherical surface of radius \( r \) gives the field of a point charge \( Q \) at the centre.
\( \displaystyle E = 0 \quad (r \lt R) \)
\( \displaystyle E = \frac{Q}{4\pi\varepsilon_0 R^2} \quad (r = R), \qquad E = \frac{Q}{4\pi\varepsilon_0 r^2} \quad (r \gt R) \)
So the graph is zero from the centre up to \( R \), jumps to its maximum at the surface, and then falls as \( 1/r^2 \). This is the graph of Option (2).
Why the other options are incorrect
Option (1) — Incorrect. It shows a constant non-zero field inside the sphere; the field inside a conductor is zero.
Option (3) — Incorrect. A field rising linearly from the centre is that of a uniformly charged non-conducting sphere, where \( E \propto r \) inside; a conductor has no field inside.
Option (4) — Incorrect. It shows the field greatest at the centre and non-zero inside; the field inside a conductor is zero.
NCERT (2026-27):
Electrostatics of Conductors – “… In the static situation, when there is no current inside or on the surface of the conductor, the electric field is zero everywhere inside the conductor.”
Electrostatics of Conductors – “… When the conductor is charged, the excess charge can reside only on the surface in the static situation.”
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.”