Identify the correct statements with regard to the application of Gauss's law in electrostatics.
(A) Gauss's law is true only for spherical closed surfaces.
(B) The Gaussian surface should not pass through any discrete charges.
(C) The total electric flux through any closed surface is zero, if no charge is enclosed by the surface.
(D) The charge in the vicinity of the surface must be zero.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (B) and (C) only
Gauss’s law states that the total electric flux through any closed surface equals the enclosed charge divided by \( \varepsilon_0 \):
\( \displaystyle \oint \vec{E}\cdot d\vec{A} = \frac{q_{\mathrm{enclosed}}}{\varepsilon_0} \)
(A) Gauss's law is true only for spherical closed surfaces. — Incorrect. It holds for a closed surface of any shape; a sphere is only a convenient choice when the charge distribution is spherically symmetric.
(B) The Gaussian surface should not pass through any discrete charges. — Correct. The field of a point charge is not defined at the charge itself, and such a charge is neither inside nor outside the surface. A Gaussian surface may, however, pass through a continuous charge distribution.
(C) The total electric flux through any closed surface is zero, if no charge is enclosed by the surface. — Correct. With \( q_{\mathrm{enclosed}} = 0 \) the flux is zero; field lines from outside charges enter and leave the surface equally, even though E itself need not be zero on the surface.
(D) The charge in the vicinity of the surface must be zero. — Incorrect. Charges outside the surface are allowed; they change the field on the surface but add nothing to the total flux.
NCERT (2026-27):
Gauss’s Law – “… (i) Gauss’s law is true for any closed surface, no matter what its shape or size.
(ii) The term \( q \) on the right side of Gauss’s law, Eq. (1.31), includes the sum of all charges enclosed by the surface. The charges may be located anywhere inside the surface.
(iii) In the situation when the surface is so chosen that there are some charges inside and some outside, the electric field [whose flux appears on the left side of Eq. (1.31)] is due to all the charges, both inside and outside \( S \). The term \( q \) on the right side of Gauss’s law, however, represents only the total charge inside \( S \).
(iv) … However, take care not to let the Gaussian surface pass through any discrete charge.”