Analysis of Each Statement

Statement A: If the electric field due to a point charge varies as ${r^{ - 2.5}}$ instead of ${r^{ - 2}},$ then the Gauss law will still be valid.

1. Gauss's Law: Gauss's law states that the total electric flux $\Phi_E$ through any closed surface (a Gaussian surface) is equal to the net charge enclosed, $q_{enc}$, divided by the permittivity of free space, $\epsilon_0$. $\Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{q_{enc}}{\epsilon_0}$
2. Dependence on Inverse Square Law: The validity of Gauss's law in this simple form is fundamentally linked to the inverse square nature of the electric field ($E \propto 1/r^2$). Let's test this. Consider a point charge $q$ at the center of a spherical Gaussian surface of radius $r$. The area of the sphere is $A = 4\pi r^2$. If $E \propto 1/r^2$, then $E = C/r^2$ for some constant $C$. The flux is: $\Phi_E = E \times A = \left( \frac{C}{r^2} \right) (4\pi r^2) = 4\pi C$ The flux is independent of the radius $r$, which is a key feature.
3. Testing the given variation: Now, let's assume the electric field varies as $E \propto r^{-2.5}$, so $E = k/r^{2.5}$. The flux through the same spherical surface would be: $\Phi_E = E \times A = \left( \frac{k}{r^{2.5}} \right) (4\pi r^2) = \frac{4\pi k}{r^{0.5}}$ In this case, the flux depends on the radius $r$ of the Gaussian surface. This means the flux is not solely dependent on the enclosed charge. Therefore, Gauss's law in its standard form would not be valid.

Conclusion: Statement A is incorrect.

Statement B: The Gauss law can be used to calculate the field distribution around an electric dipole.

1. Applicability of Gauss's Law for Calculation: While Gauss's law is always fundamentally true (the total flux is always proportional to the enclosed charge), its practical use for calculating the electric field $\vec{E}$ is limited to situations with a high degree of symmetry (spherical, cylindrical, or planar).
2. Symmetry of a Dipole: An electric dipole consists of two equal and opposite charges. The electric field it produces does not possess the simple symmetries required to use Gauss's law effectively. For any simple Gaussian surface (like a sphere), the electric field $\vec{E}$ will not be constant in magnitude nor will it be perpendicular to the surface at all points. This prevents us from simplifying the flux integral $\oint \vec{E} \cdot d\vec{A}$ to $E \oint dA$ and solving for $E$.
3. Example: If we enclose the entire dipole with a Gaussian surface, the net enclosed charge is $q_{enc} = (+q) + (-q) = 0$. So, the net flux is zero. This tells us $\oint \vec{E} \cdot d\vec{A} = 0$, but it doesn't help us find the value of $\vec{E}$ at any specific point.

Conclusion: Statement B is incorrect. Although Gauss's law is valid, it is not a useful tool for calculating the field of a dipole.

Statement C: If the electric field between two point charges is zero somewhere, then the sign of the two charges is the same.

1. Case 1: Same Sign Charges. Let two positive charges $q_1$ and $q_2$ be placed on the x-axis. At any point between them, the electric field due to $q_1$ points away from $q_1$, and the field due to $q_2$ points away from $q_2$. These two field vectors are in opposite directions. It is possible for them to cancel out. Let the distance between them be $d$. If the null point is at a distance $x$ from $q_1$, then: $E_{net} = E_1 - E_2 = \frac{k q_1}{x^2} - \frac{k q_2}{(d-x)^2} = 0$ $\frac{q_1}{x^2} = \frac{q_2}{(d-x)^2}$ This equation has a valid solution for $x$ between $0$ and $d$.
2. Case 2: Opposite Sign Charges. Let $q_1$ be positive and $q_2$ be negative. At any point between them, the field from $q_1$ points away from $q_1$ (towards $q_2$), and the field from $q_2$ points towards $q_2$. Both field vectors point in the same direction. Therefore, their sum can never be zero.

Conclusion: Statement C is correct. For the electric field to be zero at a point between two charges, the fields from the two charges must oppose each other, which only happens if the charges have the same sign.

Statement D: The work done by the external force in moving a unit positive charge from point $A$ at potential ${V_A}$ to point $B$ at potential ${V_B}$ is $\left( {{V_B} - {V_A}} \right).$

1. Definition of Electric Potential Difference: The electric potential difference between two points, $B$ and $A$, is defined as the work done per unit charge by an external agent in moving the charge from point $A$ to point $B$ without any change in kinetic energy. $\Delta V = V_B - V_A = \frac{W_{ext, A \to B}}{q}$
2. Applying to a Unit Positive Charge: The question specifies a unit positive charge, which means $q = +1$. Substituting this into the definition: $V_B - V_A = \frac{W_{ext, A \to B}}{1}$ $W_{ext, A \to B} = V_B - V_A$

Conclusion: Statement D is correct. It is the fundamental definition of potential difference.

Final Summary

• Statement A is incorrect.
• Statement B is incorrect.
• Statement C is correct.
• Statement D is correct.

The correct options are C and D.