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 an 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:

We begin with the basic definition of electric flux. By definition, the differential electric flux $$d\Phi$$ through an infinitesimal area vector $$d\vec A$$ is

$$d\Phi \;=\; \vec E \cdot d\vec A \;=\; E\,A\cos\theta,$$

where $$\theta$$ is the angle between the electric field $$\vec E$$ and the outward-drawn area vector $$d\vec A$$. Using this, we shall test each statement.

Statement (a): “The electric lines of force entering into a Gaussian surface provide negative flux.” When the field lines enter the surface, $$\theta$$ is greater than $$90^{\circ}$$, so $$\cos\theta<0$$ and $$\vec E\cdot d\vec A<0$$. Hence $$d\Phi$$ is negative. So the statement is correct.

Statement (b): “A charge $$q$$ is placed at the centre of a cube. The flux through all the faces will be the same.” By Gauss’s law, the total flux leaving the cube is

$$\Phi_{\text{total}}=\dfrac{q}{\varepsilon_{0}}.$$

Because the charge is exactly at the centre, the cube is perfectly symmetric; therefore the flux is distributed equally among all six faces. Hence the flux through one face is

$$\Phi_{\text{one face}}=\dfrac{1}{6}\left(\dfrac{q}{\varepsilon_{0}}\right).$$

This shows that each face indeed receives the same flux, so statement (b) is correct.

Statement (c): “In a uniform electric field, net flux through a closed Gaussian surface containing no net charge is zero.” Gauss’s law states

$$\oint\vec E\cdot d\vec A=\dfrac{q_{\text{enc}}}{\varepsilon_{0}}.$$

If the surface encloses no charge, then $$q_{\text{enc}}=0$$, and therefore

$$\oint\vec E\cdot d\vec A = 0.$$

This is true regardless of whether the external field is uniform or not, so the statement is certainly correct.

Statement (d): “When an electric field is parallel to a Gaussian surface, it provides a finite non-zero flux.” If the field is exactly parallel to the surface, the angle $$\theta$$ between $$\vec E$$ and the outward normal $$d\vec A$$ is $$90^{\circ}$$. Hence

$$\cos\theta=\cos90^{\circ}=0,$$

so each elemental flux $$d\Phi=E\,A\cos\theta=0.$$ Integrating over the whole surface, the total flux also remains zero. Therefore the flux is not finite and non-zero; it is strictly zero. Thus statement (d) is incorrect.

We have found that statements (a), (b) and (c) are correct, while statement (d) is incorrect. Among the given options, only Option D states “(d) Only”.

Hence, the correct answer is Option D.