Question

Choose the incorrect statement :
(a) The electric lines of force entering into a Gaussian surface provide negative flux.
(b) A charge 'q' is placed at the centre of a cube. The flux through all the faces will be the same.
(c) In a uniform electric field net flux through a closed Gaussian surface containing no net charge, is zero.
(d) When electric field is parallel to a Gaussian surface, it provides a finite non-zero flux.
Choose the most appropriate answer from the options given below :

• (a) and (c) Only
• (b) and (d) Only
• (c) and (d) Only
• (d) Only

Hint:

Always remember: Electric flux is a measure of the number of field lines "piercing" a surface. If the field is parallel to the surface, no lines pierce it, so flux must be zero.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Electric flux (\(\Phi\)) through a surface is given by the surface integral of the electric field over that surface: \(\Phi = \oint \vec{E} \cdot d\vec{A}\). According to Gauss's Law, the net flux through a closed surface depends only on the net charge enclosed.
Step 2: Detailed Explanation:
(a) By convention, field lines entering a surface represent negative flux, and those leaving represent positive flux. This is correct.
(b) Due to symmetry, if a charge is at the center of a cube, the flux through each of the 6 faces is exactly \(q/6\epsilon_0\). This is correct.
(c) According to Gauss's Law, if \(q_{enclosed} = 0\), the net flux \(\Phi = 0\), regardless of whether the external field is uniform or not. This is correct.
(d) If the electric field \(\vec{E}\) is parallel to the surface, it is perpendicular to the area vector \(d\vec{A}\) (which is always normal to the surface). Thus, \(\vec{E} \cdot d\vec{A} = E dA \cos 90^\circ = 0\). The flux is zero, not a finite non-zero value. This is incorrect.
Step 3: Final Answer:
Only statement (d) is incorrect.