Start from Gauss's law

$\oint \vec E \cdot d\vec A = \frac{q_{\text{enc}}}{\varepsilon_0}$

When can this become

$|\vec E|=\frac{q_{\text{enc}}}{\varepsilon_0 |A|}?$

For this simplification, the flux integral must reduce to

$\oint \vec E \cdot d\vec A = E \oint dA = E A$

This is possible only if:

• the magnitude $E$ is constant over the whole Gaussian surface, and
• $\vec E$ is parallel to $d\vec A$ everywhere, so that $\vec E \cdot d\vec A = E\,dA$

The second condition means the electric field is normal to the surface everywhere.

Relation with equipotential surface

Electric field is always perpendicular to an equipotential surface. Hence, if the Gaussian surface is an equipotential surface, then $\vec E$ is normal to the surface everywhere.

So to write

$\oint \vec E \cdot d\vec A = EA$

we need both:

• Gaussian surface is an equipotential surface (to ensure direction is normal everywhere), and
• $|\vec E|$ is constant on the surface.

Check options

A: Only when $|\vec E|$ is constant on the surface.
Not sufficient, because direction of $\vec E$ relative to $d\vec A$ may vary. So A is false.

B: For any choice of Gaussian surface.
False. Gauss's law is always true, but this simplified formula is not valid for arbitrary surfaces. So B is false.

C: Only when the Gaussian surface is an equipotential surface.
Not sufficient, because even on an equipotential surface, $E$ may vary from point to point. So C is false.

D: Only when the Gaussian surface is an equipotential surface and $|\vec E|$ is constant on the surface.
This gives both required conditions. So D is correct.

Final answer

$\boxed{D}$