σ is the uniform surface charge density of a thin spherical shell of radius R . The electric field at any point on the surface of the spherical shell is :
(A)σ/ϵ0R
(B)σ/2∈0
(C)σ/ϵ0✓ Correct
(D)σ/4ϵ0
Answer: (C)
Solution
By Gauss's law, just outside the shell E⋅4πR2=ϵ0σ⋅4πR2, giving E=ϵ0σ (the field is zero just inside). The field at the surface of the shell is taken as
E=ϵ0σ.
JEE Main · 2024 · 06 Apr 2024 · Shift 1 · Single correct option