Electric Flux and Electric Dipole

Zero net charge means the algebraic sum of the dipole’s charges is zero; it does not mean that the forces on the separated charges cancel in every situation. In a non-uniform field, the positive and negative charges are at different locations and may experience different field strengths. Consequently, the attractive and repulsive forces have unequal magnitudes, leaving a resultant force that can pull the dipole toward the stronger-field region. Option A is correct.

For a point far from a short electric dipole on its axial line, the exact field expression reduces approximately to Eaxial = (1/4πε₀)(2p/r³). Therefore the field varies as 1/r³, not as 1/r or 1/r². Replacing r by 2r gives E′/E = r³/(2r)³ = 1/8. Hence the new field is one eighth of the original, so option A is correct.

At a far point on the equatorial line of a short dipole, the field magnitude is approximately Eequatorial = (1/4πε₀)(p/r³). The direction is opposite to the dipole moment, but its distance dependence is still 1/r³. If r changes to 3r, then E′/E = 1/(3³) = 1/27. Therefore one twenty-seventh of the original field remains, making option A correct.

For a short dipole at a distant point, the axial field magnitude is Eₐ = (1/4πε₀)(2p/r³), while the equatorial field magnitude is Eₑ = (1/4πε₀)(p/r³). At the same distance and for the same dipole moment, the common factors cancel, giving Eₐ/Eₑ = 2. Their directions differ, but the question asks for magnitudes; therefore option A is correct.

Place the dipole on a horizontal axis with the positive charge on the right and the negative charge on the left; the dipole moment points from negative to positive. At an equatorial point, the distances from both charges are equal, so the components perpendicular to the dipole axis cancel by symmetry. The components along the axis point toward the negative charge and add. Thus the resultant field is opposite to p, so option A is correct.

A point charge produces a field whose leading dependence is E ∝ 1/r². A dipole consists of equal and opposite charges separated by a small distance. At a far point, the two 1/r² contributions nearly cancel because of their opposite signs; the remaining term depends on the separation and falls as E ∝ p/r³. Thus the dipole field decreases faster with distance. The other options deny charge or field properties that are essential.

The far axial electric field of a dipole has the proportionality E_axial ∝ p/r³. Therefore, the changed-to-original ratio is E′/E = (2p/p) × (r/2r)³ = 2 × (1/2)³ = 2 × 1/8 = 1/4. The doubled dipole moment increases the field by two, but doubling distance decreases it by eight, so the net field is one quarter. The other choices ignore one of these factors.

For a far dipole field, E ∝ p/r³. If p changes to 3p and r changes to 3r, then E′/E = 3/(3³) = 3/27 = 1/9. Thus one ninth of the original far field remains. One third would result from using an inverse-first-power dependence, while three times would ignore the distance effect and unchanged would ignore both changes. The inverse-cube dependence is the governing concept.

For a point sufficiently far from a short electric dipole, the axial electric-field magnitude varies as E ∝ 1/r³. If the distance changes from r to 2r, the new field is E′ = E(r/2r)³ = E/8. With the original magnitude E = 40, E′ = 40/8 = 5. Therefore option A is correct. The values 10 and 20 would correspond to inverse-square or inverse-first-power behavior, not dipole-field dependence.

For a short electric dipole at the same far distance, the axial field magnitude is E axial = 2kp/r³, while the equatorial field magnitude is E equatorial = kp/r³. Thus E axial = 2E equatorial. Given the equatorial field is 20, the axial field is 2 × 20 = 40. Therefore option A is correct. The factor of two is the characteristic axial-to-equatorial ratio, so the other numerical choices do not follow the dipole-field expressions.

Consider a point on the perpendicular bisector, or equatorial line, of an electric dipole. The point is equally distant from both charges, so the components of the two electric fields parallel to the equatorial direction are equal and opposite and cancel. Their components along the dipole axis point in the same direction, which is opposite to the dipole moment defined from negative to positive charge. Thus the resultant equatorial field is opposite to p, making option A correct.

For a closed surface, Gauss’s law connects total flux with net enclosed charge, not directly with dipole moment. A dipole contains charges +q and −q, so its net charge is zero and its total flux is zero. However, the dipole moment is p = qd, where d is the separation vector between the charges. It can therefore be non-zero even when the net charge and total flux are zero. Hence option A is correct.

A point charge produces an electric field proportional to 1/r². A dipole has equal and opposite charges separated by a small distance, so at a far observation point the two nearly equal fields oppose one another and largely cancel. The remaining dipole field is proportional to p/r³ in the far-field approximation, where p is the dipole moment. Thus it decreases faster with distance. The cancellation, not absence of a negative-charge field, explains the result.

In the far-field approximation, the axial field magnitude of a dipole is Eaxial = (1/4πε0)(2p/r³), while the equatorial field magnitude is Eequatorial = (1/4πε0)(p/r³). Their numerical magnitudes differ by a factor of two and their directions are generally different, but both vary as 1/r³. Therefore option A states the common property. Options B and C ignore direction and the factor-of-two difference, while D is false.

Gauss’s law gives the total flux through a closed surface as Φ = Q_enclosed/ε₀. A complete electric dipole contains equal charges +q and −q, so its net enclosed charge is q − q = 0. Hence Φ = 0, irrespective of the size or shape of the closed surface, provided the entire dipole remains enclosed. Option A is correct; the flux is not determined by dipole moment, surface shape, or an always-negative sign.

For any closed surface, Gauss’s law states Φ = Q_enclosed/ε₀. In this situation the surface encloses only the negative charge, say −q, while the positive charge lies outside and is not included in Q_enclosed. Thus Φ = −q/ε₀, so the total flux is negative. Option A is correct. It is not zero because the enclosed charge is not balanced, and it does not depend on surface area for a fixed enclosed charge.

Gauss’s law relates the net flux through a closed surface to the algebraic sum of charges inside it: Φ = Q_net/ε₀. If Φ = 0, then Q_net = 0. This does not mean that the electric field is zero at every point; fields from several charges can cancel in total flux. Equal positive and negative charges may also be enclosed. Therefore option A is the safest conclusion, while B, C, and D make claims not supported by the given information.

Electric flux through a surface is Φ = E A cos θ for a uniform field over a plane surface, where θ is the angle between the field and the area vector. If the field is parallel to the surface, it is perpendicular to the area vector, so θ = 90° and cos θ = 0; flux is then zero even though E is non-zero. Thus option A is correct. Zero flux measures the normal component, not necessarily the entire field.

Gauss’s law says that total flux through a closed surface depends on the net charge enclosed, not simply on every field present at the surface. Field lines from an external charge can enter the closed surface at some locations and leave it at others. Their inward and outward contributions have opposite signs and cancel in the algebraic total, so the external charge contributes zero net closed flux. Option A is correct; it does create a field, and flux also exists for closed surfaces.

For a closed surface, Gauss’s law gives Φ_total = Q_net/ε₀. A field entering through one portion produces negative local flux there, but that does not determine the total. The total is the algebraic sum over every part of the surface. Since the net enclosed charge is positive, Q_net > 0 and therefore Φ_total > 0. Option A is correct. Local inward flux may coexist with positive total flux; options B, C, and D ignore the complete surface or the enclosed-charge rule.

The governing concept is Gauss’s law: the net electric flux through a closed surface equals the enclosed net charge divided by ε₀. A complete dipole contains equal positive and negative charges, so its enclosed charge is zero. Local flux is not zero: outward field contributes positive flux and inward field contributes negative flux. These signed contributions cancel in the surface integral, making the total flux zero. Thus A is correct; B, C, and D confuse local field, dipole moment, or surface shape with net flux.

A dipole consists of equal and opposite charges separated by a small distance. At a far point, the two individual Coulomb fields are nearly equal in magnitude and opposite in direction, so their leading 1/r² contributions largely cancel. The remaining dipole term decreases approximately as 1/r³, faster than the 1/r² field of a single point charge. Thus A explains the different falloff; B, C, and D are physically incorrect.

For a short dipole at distance r, the far axial field magnitude is E_axial = (1/4πε₀)(2p/r³), whereas the far equatorial field magnitude is E_equatorial = (1/4πε₀)(p/r³). At the same distance, the common factors cancel, leaving E_axial/E_equatorial = 2. Therefore the axial field is twice the equatorial field, so A is correct. The fields are not equal or zero for a nonzero dipole.

Take the dipole moment from the negative charge toward the positive charge and consider a point on the perpendicular bisector. By symmetry, the components of the two charge fields perpendicular to the dipole axis cancel. Their components along the axis point in the same direction, but that direction is opposite to the dipole moment. Consequently the equatorial field is E = −(1/4πε₀)p/r³ in the far-field limit. Hence A is correct.

The dipole moment points from the negative charge toward the positive charge. On a far point along the axial line beyond the positive charge, the field due to the positive charge points away from it, along the dipole moment. The field due to the negative charge points toward the negative charge; its axial direction at that point is also along the dipole moment after the vector geometry is considered. The components therefore reinforce, giving E_axial ≈ (1/4πε₀)2p/r³. Thus A is correct.