In the expression of Gauss's Law.
\( \int\vec{E}\cdot d\vec{s}=q/\varepsilon_0 \)
(A) The electric field on LHS is only due to the charges inside the Gaussian surface.
(B) The electric field on LHS is due to charges inside and outside the Gaussian surface.
(C) The electric charge on RHS represents all the charges inside and outside the Gaussian surface.
(D) The electric charges on RHS represent only the charges outside the Gaussian surface.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (B) only
In Gauss’s law \( \oint\vec{E}\cdot d\vec{s} = q/\varepsilon_0 \), the field \( \vec{E} \) on the left is the total field at each point of the surface, produced by all charges. The charge q on the right is only the charge enclosed by the surface.
(A) The electric field on LHS is only due to the charges inside the Gaussian surface. — Incorrect. \( \vec{E} \) is due to charges both inside and outside the surface, not just inside.
(B) The electric field on LHS is due to charges inside and outside the Gaussian surface. — Correct. Outside charges contribute to \( \vec{E} \) at the surface, but their net flux through the closed surface is zero, so they do not appear in q.
(C) The electric charge on RHS represents all the charges inside and outside the Gaussian surface. — Incorrect. q counts only the enclosed charge.
(D) The electric charges on RHS represent only the charges outside the Gaussian surface. — Incorrect. Charges outside the surface do not contribute to q at all.
NCERT (2026-27):
Gauss’s Law – “… (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.”