A charge of \( 50\,\mu\mathrm{C} \) is uniformly distributed on the surface of a thin spherical shell of radius 10 cm. The electric field at a point at a distance of 5 cm from the center of the shell is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Zero
By Gauss's law, the electric field at a point depends only on the charge enclosed by a spherical Gaussian surface through that point.
The point is at \( r = 5\,\mathrm{cm} \), inside the shell of radius \( 10\,\mathrm{cm} \). A Gaussian sphere of radius 5 cm encloses no charge, because all the charge sits on the surface of the shell.
\( \displaystyle E\,(4\pi r^{2}) = \frac{q_{\mathrm{enclosed}}}{\varepsilon_0} = 0 \;\Rightarrow\; E = 0 \)
So the field inside a uniformly charged thin shell is zero at every point.
NCERT (2026-27):
Field Due to a Uniformly Charged Thin Spherical Shell – “… (ii) Field inside the shell: In Fig. 1.28(b), the point P is inside the 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.”