Electric Flux and Electric Dipole

Gauss’s law gives the net flux through any closed surface as Φ = Q_enclosed/ε₀. The shape of the Gaussian surface does not appear in this relation. Changing a sphere into an irregular surface may alter the field strength, surface orientation and local flux contributions, but their total remains fixed when the enclosed net charge is unchanged. Therefore option A is correct; the field need not be uniform and irregular surfaces can certainly have flux.

A complete dipole contains equal positive and negative charges, so its net enclosed charge is zero. Gauss’s law therefore gives zero net flux through the closed surface. But zero net flux is an integral result: positive and negative contributions can cancel. Because the two charges are separated, they generally produce a nonzero electric field at individual points on the surface. Hence option A is correct; zero flux does not imply zero field everywhere.

Gauss’s law states that the net flux through a closed surface is Φ = Q_enclosed/ε₀. Initially, only +q is enclosed, so the flux is +q/ε₀ and is positive. After enlargement, both +q and −q are enclosed; their algebraic sum is zero. Therefore Φ becomes zero. The surface shape or size alone is not decisive; only the net enclosed charge matters.

Flux is defined by the dot product Φ = ∫ E·dA. For an open surface, there is no unique outward normal, so one must specify which of the two possible normal directions is chosen for dA. Reversing that choice reverses the sign of the flux, although its magnitude is unchanged. Thus the area-vector convention is essential for an unambiguous answer.

A closed surface has a well-defined outward normal at every small element. With this convention, field crossing outward gives E·dA > 0, while field entering the surface gives E·dA < 0. This consistent sign convention is what makes Gauss’s law, ∮E·dA = Q_enclosed/ε₀, easy to apply. It changes no field, charge, or geometry; it only defines orientation.

For a small surface element, flux is dΦ = E·dA = E A cos θ. The field can be resolved into a normal component E⊥ and a tangential component E∥. The normal component crosses the surface and contributes E⊥A, whereas the tangential component runs along the surface and contributes zero to the dot product. Therefore only E⊥ determines the flux.

For a uniform field through a plane surface, Φ = EA cos θ. Initially Φ₁ = EA cos 30° = (√3/2)EA. Finally Φ₂ = EA cos 60° = (1/2)EA. Therefore Φ₂/Φ₁ = (1/2)/(√3/2) = 1/√3, so the flux decreases to about 0.577 of its original value. It does not become zero because 60° is not 90°.

Electric field direction and dipole-moment direction use different conventions. Electric field lines outside a dipole point from the positive charge toward the negative charge, so at the centre the field is directed positive to negative. By definition, however, the electric dipole moment p is the vector from the negative charge to the positive charge, with magnitude p = qd. Therefore option A is correct, not option B.

The magnitude of an electric dipole moment is p = qd, where q is the magnitude of either charge and d is the separation between the charges. Initially p = qd. After the changes, q′ = 2q and d′ = 2d, so p′ = q′d′ = (2q)(2d) = 4qd = 4p. The direction remains from the negative charge toward the positive charge if the charge arrangement is unchanged. Hence option A is correct.

The electric dipole moment has magnitude p = qd, where q is the magnitude of either charge and d is the separation between the charges. Initially p = qd. After the change, q' = q/2 and d' = 4d, so p' = (q/2)(4d) = 2qd = 2p. Therefore, the dipole moment becomes double. It is not merely half or four times because both changes must be included.

An electric dipole consists of equal and opposite charges separated by a finite distance. Their algebraic sum is zero, so the net charge is zero, but the dipole moment is a vector quantity p = qd directed from the negative charge to the positive charge. Because d is non-zero, p can also be non-zero. Option B contradicts the definition, option C removes the separation, and option D confuses net charge with dipole moment.

A permanent dipole moment requires a stable separation between the centres of positive and negative charge. In a non-polar molecule, the charge distribution is symmetric, so these effective centres coincide and the vector sum of molecular dipoles is zero. An external field may induce a temporary dipole, but that does not create a permanent one. Options B, C and D do not describe non-polar molecules.

Electric flux through a plane surface is Φ = EA cos θ, where θ is the angle between the electric field and the area vector, which is normal to the surface. If the field is oblique, only its normal component E cos θ contributes. A common error is to use Φ = EA and ignore the angle, thereby counting the full field rather than its perpendicular component. Including area, units, and an area-vector direction are necessary parts of a correct calculation.

Electric flux through a plane surface is given by Φ = EA cos θ, so it depends multiplicatively on the electric field, area, and angular factor cos θ. The changes produce the ratio Φ′/Φ = 3 × 2 × 1/2 = 3. Therefore, the new flux is three times the original flux. Six times would incorrectly omit the halving factor, while half and unchanged are inconsistent with the product.

For a uniform field crossing a plane surface, Φ = EA cos θ, where θ is the angle between the electric field and the area vector. When the area vector is along the field, θ = 0° and cos θ = 1, its greatest possible value. Thus Φ = EA, meaning the entire field component contributes through the surface. If the field ran along the surface, θ would be 90° and flux would be zero.

Gauss’s law states that the net flux through a closed surface is Φ = Q_enclosed/ε₀. Thus the first step is to determine the algebraic net charge actually inside the surface. A zero flux does not imply zero electric field: a dipole has zero net charge but can produce a nonzero local field. Hence A separates global flux from local field; B, C, and D make unjustified assumptions.

For a plane surface in a uniform field, electric flux is Φ = E A cos θ, where θ is measured between the electric field and the area vector. Initially θ = 60°, so Φ₁ = EA cos 60° = EA/2. After rotation they are parallel, θ = 0°, giving Φ₂ = EA. Therefore Φ₂/Φ₁ = EA/(EA/2) = 2, so A is correct; B reverses the comparison.

The magnitude of a dipole moment is p = qd, where q is the magnitude of either charge and d is the separation; its direction is from negative to positive charge. If q becomes 3q and d becomes d/2, the new moment is p′ = (3q)(d/2) = 3qd/2 = 1.5p. Therefore A is correct. B ignores the reduced separation, C ignores the charge increase, and D multiplies both changes incorrectly.

Dipole moment has magnitude p = qd. To preserve the initial moment, the product of charge magnitude and separation must remain constant: qd = q′d′. With q′ = 4q, we require qd = 4qd′, so d′ = d/4. Thus the separation must become one-fourth and A is correct. Making it four times would increase the moment, halving it would still double the moment, and leaving it unchanged would quadruple the moment.

Electric flux through a plane surface is Φ = EA cos θ, where θ is the angle between the electric field and the chosen area vector, which is normal to the surface. A positive flux means the cosine is positive; a negative flux means the relative orientation has crossed to a negative cosine. Rotating the surface changes the area-vector direction relative to the field, not the field unit, mass, or charge nature. Hence option A is correct.

At the same distance from a short electric dipole, the magnitude of the axial field is twice the magnitude of the equatorial field: E axial = 2E equatorial. The axial field is given as 60, so E equatorial = 60/2 = 30. Therefore option A is correct. The answer is not 60 because the two fields are not equal, and it is not 120 because the equatorial field is smaller rather than larger than the axial field.

The governing idea is the vector addition of the fields produced by the positive and negative charges. On the axial line, both individual fields lie along the same line, and at a far point their resultant has the direction from the negative charge toward the positive charge, which is the direction of dipole moment. Therefore A is correct; B, C, and D incorrectly ignore or reverse the charge contributions.

Zero net charge means only that the algebraic sum of the two charges is zero; it does not mean that the charges occupy the same position. A dipole has separated positive and negative charges, so their fields generally do not cancel at every observation point. At large distance the dipole field decreases approximately as 1/r³, but it remains finite and non-zero in general. Therefore A is correct.

The electric field of an isolated point charge varies as E_point ∝ 1/r². The far field of an electric dipole varies as E_dipole ∝ 1/r³. When distance is doubled, the point-charge field becomes one-fourth, whereas the dipole field becomes one-eighth. Since one-eighth is a greater reduction than one-fourth, the dipole field decreases more rapidly. Hence A is correct.

Electric flux through a plane surface is given by Φ = EA cosθ. Since the angle remains unchanged, cosθ is constant. The field becomes 4E and the area becomes A/2, so the new flux is (4E)(A/2)cosθ = 2EA cosθ = 2Φ. Therefore, the flux becomes double. Option B is incorrect because it considers only the halved area, while option C considers only the increased field.