Which of the following statement(s) is/are correct?
(a) If the electric field due to a point charge

Q: Why exactly does Gauss's law fail if the force is not inverse square?

Gauss's law relies on the fact that the surface area of a sphere grows as $r^2$, while the electric field weakens as $1/r^2$. These two exactly cancel out, making the total flux independent of the radius. If the field varied as $r^{-2.5}$, the flux would depend on the size of the surface, breaking the law.

Q: But isn't Gauss's law always true? Why can't we use it for a dipole?

Gauss's law *is* always true for any closed surface! However, the statement says 'used to calculate the field distribution'. To actually calculate $\vec{E}$, we need to pull it out of the integral $\oint \vec{E} \cdot d\vec{A}$. This requires high symmetry (like a sphere or cylinder), which a dipole lacks.

Q: Can a null point exist between two opposite charges if they have different magnitudes?

No. Between two opposite charges, the electric field vectors from both charges will always point in the same direction (away from the positive and towards the negative). They can never cancel each other out in that region, regardless of their magnitudes. The null point would exist *outside* the region between them.

Q: What is the difference between work done by an external force and work done by the electric field?

The work done by an external force (moving a charge slowly without changing kinetic energy) is equal to the positive change in potential energy: $W_{ext} = \Delta U = q(V_B - V_A)$. The work done by the conservative electric field is the negative of the change in potential energy: $W_{elec} = -\Delta U = q(V_A - V_B)$.

Q: Why do we specify a 'unit positive charge' in statement (d)?

The electric potential difference $V_B - V_A$ is defined as the work done per unit charge. By specifying a unit positive charge ($q = +1$), the work done $W = q(V_B - V_A)$ numerically equals the potential difference itself.

Question

Which of the following statement(s) is/are correct?

Options:

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

(B) The Gauss's law can be used to calculate the field distribution around an electric dipole

(C) If the electric field between two point charges is zero somewhere, then the sign of the two charges is the same

(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 $(V_B - V_A)$

Sigma Percentile · Accepted Answer

Welcome to an exciting exploration of electrostatics! In this problem, we are tasked with evaluating four distinct statements that test our fundamental understanding of Gauss's law, electric fields, and electrostatic potential. Let's break down each statement and uncover the physics behind it.

### Analyzing Statement A: Gauss's Law and the Inverse Square Law

The first statement claims that Gauss's law would still be valid even if the electric field due to a point charge varied as $r^{-2.5}$ instead of the familiar $r^{-2}$. To see why this is incorrect, we must look at the mathematical heart of Gauss's law.

Gauss's law states that the total electric flux through a closed surface is proportional to the enclosed charge:

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

This elegant relationship is not a coincidence; it is a direct mathematical consequence of the inverse square law. When we calculate the flux through a spherical Gaussian surface of radius $r$, the surface area grows as $4\pi r^2$. If the electric field weakens exactly as $1/r^2$, the $r^2$ terms perfectly cancel out, leaving the total flux completely independent of the sphere's radius.

However, if the electric field varied as $r^{-2.5}$, the flux would become dependent on the radius $r$. The flux would change as we expand or shrink our Gaussian surface, completely breaking the fundamental premise of Gauss's law. Therefore, statement (a) is incorrect.

### Analyzing Statement B: Symmetry and Gauss's Law

The second statement suggests that Gauss's law can be used to calculate the field distribution around an electric dipole.

It is crucial to distinguish between Gauss's law being *valid* and being *useful*. Gauss's law is universally valid for any closed surface. However, to actually calculate the electric field $\vec{E}$, we need to pull the electric field magnitude out of the flux integral. This mathematical maneuver is only possible when the charge distribution possesses a high degree of spatial symmetry—such as perfect spherical, cylindrical, or planar symmetry.

An electric dipole, consisting of two equal and opposite charges separated by a small distance, lacks this required symmetry. The electric field magnitude is not constant over any simple closed surface we could draw around it. Consequently, we cannot easily use Gauss's law to find its field distribution. Statement (b) is also incorrect.

### Analyzing Statement C: The Null Point

The third statement proposes that if the electric field between two point charges is zero somewhere, then the sign of the two charges must be the same.

Imagine a point $P$ located on the line segment joining two charges. For the net electric field to be zero at $P$, the individual electric field vectors created by each charge must perfectly cancel each other out. This means they must be equal in magnitude but point in exactly **opposite directions**.

$$ \vec{E}_{net} = \vec{E}_1 + \vec{E}_2 = 0 \implies \vec{E}_1 = -\vec{E}_2 $$

If the two charges had opposite signs (one positive, one negative), their electric fields in the region between them would point in the *same* direction—away from the positive charge and towards the negative charge. They would add up, not cancel! Therefore, to achieve a null point between them, both charges must have the same sign (either both positive or both negative) so that their fields oppose each other. Statement (c) is absolutely correct.

### Analyzing Statement D: Work and Potential Energy

The final statement defines the work done by an external force in moving a unit positive charge from point $A$ at potential $V_A$ to point $B$ at potential $V_B$ as $(V_B - V_A)$.

By definition, the work done by an external force in moving a charge $q$ slowly (without changing its kinetic energy) against an electrostatic field is equal to the change in the system's electrostatic potential energy:

$$ W_{ext} = \Delta U = q(V_B - V_A) $$

The statement specifically asks about a **unit positive charge**, which means we substitute $q = +1$.

$$ W_{ext} = (+1)(V_B - V_A) = V_B - V_A $$

This perfectly matches the expression given in the statement. Thus, statement (d) is also correct.

### Conclusion

After a thorough conceptual analysis, we find that statements (c) and (d) are the correct choices. This problem beautifully highlights why we must never memorize formulas blindly, but rather understand the geometric and physical constraints that make them true!

Animated Solution for Physics - Electrostatics: Which of the following statement(s) is/are correct?

Select Answer:

Visualized Solution

Analyzing the Statements

Gauss's Law and Inverse Square Law

Invalidating Option (a)

Gauss's Law for a Dipole

Invalidating Option (b)

Null Point Between Two Charges

Validating Option (c)

Work Done by External Force

Raw Setup (Substitution)

Validating Option (d)

Final Answer

The Way Forward

The Sigma Insight: Electric Field Lines, Flux and Gauss's Law

Analyzing Statement A

Analyzing Statement B

Analyzing Statement C

Analyzing Statement D

Conclusion

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