Validity of Gauss Law Electric Field and Potential Properties

To determine which statements are correct, let us analyze each option individually:

1. Analysis of Statement (A):
   Gauss's law states that the net electric flux through any closed Gaussian surface depends solely on the enclosed charge:
   $\Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{q_{\text{enc}}}{\varepsilon_0}$
   For a point charge $q$ inside a spherical Gaussian surface of radius $r$, the area is $A = 4\pi r^2$.

   • If $E \propto r^{-2}$, then:
     $\Phi_E \propto (r^{-2})(4\pi r^2) = \text{constant (independent of } r \text{)}$
   • If $E \propto r^{-2.5}$, then:
     $\Phi_E \propto (r^{-2.5})(4\pi r^2) \propto r^{-0.5}$
     Since the electric flux through a closed surface enclosing a charge would depend on the distance $r$ from the charge, Gauss's law in its standard formulation would no longer be valid.
     Thus, Statement (A) is incorrect.

2. Analysis of Statement (B):
   While Gauss's law is always fundamentally true, it can only be used to explicitly calculate the electric field if the charge distribution exhibits a high degree of spatial symmetry (such as spherical, cylindrical, or planar symmetry). An electric dipole does not possess sufficient symmetry for the electric field to be factored out of the surface integral $\oint \vec{E} \cdot d\vec{A}$.
   Thus, Statement (B) is incorrect.

3. Analysis of Statement (C):
   Consider two point charges $q_1$ and $q_2$ located on the $x$-axis at $x = 0$ and $x = d$ respectively.
   At a point $P$ located between the two charges ($0 < x < d$), the electric field contributed by $q_1$ is directed along $\hat{i}$ (if $q_1 > 0$), while the electric field contributed by $q_2$ is directed along $-\hat{i}$ (if $q_2 > 0$).

   The net electric field at point $P$ is:
   $\vec{E} = \frac{1}{4\pi\varepsilon_0} \left[ \frac{q_1}{x^2} \hat{i} - \frac{q_2}{(d - x)^2} \hat{i} \right]$
   Setting $\vec{E} = 0$ gives:
   $\frac{q_1}{x^2} = \frac{q_2}{(d - x)^2}$
   Since $x^2 > 0$ and $(d - x)^2 > 0$, $q_1$ and $q_2$ must have the same algebraic sign ($q_1 q_2 > 0$) for this equality to hold. If the charges had opposite signs, the electric fields produced by both charges between them would point in the same direction, making a zero field impossible between them.
   Thus, Statement (C) is correct.

4. Analysis of Statement (D):
   The potential difference $V_B - V_A$ between two points $B$ and $A$ in an electrostatic field is defined as the work done by an external force ($W_{\text{ext}}$) in slowly moving a test charge $q$ from point $A$ to point $B$:
   $W_{\text{ext}} = q (V_B - V_A)$
   For a unit positive charge ($q = +1$), this yields:
   $W_{\text{ext}} = V_B - V_A$
   Thus, Statement (D) is correct.

Correct Answer: C and D