What is the electric field at a point outside a uniformly charged thin spherical shell of total charge q, at a distance r from its centre?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \frac{q}{4\pi\varepsilon_0 r^{2}} \)
By Gauss’s law, the field at a point outside a uniformly charged thin spherical shell is radial. The whole charge acts as if concentrated at the centre. A spherical Gaussian surface of radius \( r \) encloses the full charge \( q \):
\( \displaystyle E\times4\pi r^{2}=\frac{q}{\varepsilon_0} \)
\( \displaystyle \Rightarrow\; E=\frac{q}{4\pi\varepsilon_0 r^{2}} \)
Option (1) \( \frac{q}{4\varepsilon_0 r^{2}} \) — Incorrect. This leaves out the factor \( \pi \). The area of a sphere is \( 4\pi r^{2} \), so \( \pi \) must appear in the field.
Option (2) \( \frac{q}{4\pi\varepsilon_0 r^{2}} \) — Correct. Gauss’s law over the sphere of area \( 4\pi r^{2} \) gives this field. It is the same as the field of a point charge \( q \) at the centre.
Option (3) \( \frac{q}{8\varepsilon_0 r^{2}} \) — Incorrect. It has no \( \pi \), and the factor 8 does not come from the area \( 4\pi r^{2} \) of the Gaussian surface.
Option (4) \( \frac{q}{8\pi\varepsilon_0 r^{2}} \) — Incorrect. This is half the correct field. The shell’s whole charge \( q \) is enclosed, so no factor of one half appears.
NCERT (2026-27):
Field Due to a Uniformly Charged Thin Spherical Shell – “… Vectorially,
\( \displaystyle \mathbf{E}=\frac{q}{4\pi\varepsilon_0 r^{2}}\,\hat{\mathbf{r}} \)\( \qquad (1.34) \)
The electric field is directed outward if q > 0 and inward if q < 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.”