Question

State Gauss' law in electrostatics. Charge inside a Gaussian surface is zero. Comment on the electric field intensity at each point of the surface.

Hint:

Net flux depends only on enclosed charge, but the local field can be produced by charges lying outside the surface. Zero flux is not zero field.

Answer 1

Step 1: State Gauss' law.
Gauss' law states that the net electric flux through any closed (Gaussian) surface equals the net charge enclosed divided by the permittivity of free space:
\[\oint \vec{E}\cdot d\vec{A}=\frac{q_{enc}}{\varepsilon_0}\]Step 2: Put the enclosed charge equal to zero.
Given \(q_{enc}=0\), so
\[\oint \vec{E}\cdot d\vec{A}=\frac{0}{\varepsilon_0}=0\]The net flux through the surface is zero.
Step 3: Comment on the field at each point.
Zero net flux does not mean the electric field is zero at every point of the surface. It only means that the total inward flux equals the total outward flux. If there are charges outside the Gaussian surface, they can still produce a non-zero field \(\vec{E}\) at individual points of the surface; the field lines that enter the surface also leave it, so their contributions cancel in the surface integral.
Conclusion: Net flux is zero, but \(\vec{E}\) at each point of the surface may be non-zero (it is zero only in the special case when no charges exist anywhere).
\[\boxed{\Phi_{net}=0,\ \text{but } \vec{E}\ \text{need not be zero on the surface}}\]