An atom has a positively charged point nucleus of charge Ze surrounded by a uniform density of negative charge upto a radius R. The atom is taken as neutral. The electric field at a distance r < R from the nucleus is given by-
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{Ze}{4\pi\varepsilon_0}\left[\frac{1}{r^2}-\frac{r}{R^3}\right] \)
The charge distribution is spherically symmetric, so use Gauss’s law with a spherical Gaussian surface of radius r. The atom is neutral, so the negative charge −Ze is spread uniformly over the sphere of radius R.
\( \displaystyle \rho = \frac{-Ze}{\frac{4}{3}\pi R^{3}} \)
Charge enclosed within radius r: the nucleus plus the negative charge inside r.
\( \displaystyle q_{\mathrm{enc}} = Ze + \rho\cdot\frac{4}{3}\pi r^{3} = Ze\left(1 - \frac{r^{3}}{R^{3}}\right) \)
\( \displaystyle E\cdot4\pi r^{2} = \frac{q_{\mathrm{enc}}}{\varepsilon_0} \)
\( \displaystyle E = \frac{Ze}{4\pi\varepsilon_0}\left[\frac{1}{r^{2}} - \frac{r}{R^{3}}\right] \)
As a check, at r = R the field is zero, as it must be outside a neutral atom.
Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27):
Field Due to a Uniformly Charged Thin Spherical Shell – …
