Electric Flux and Electric Dipole

The governing idea is the direction of the electric field produced by a dipole. The dipole moment points from the negative charge to the positive charge. On the axial line, the fields combine in the direction of the dipole moment. On the equatorial line, the components combine opposite to that moment. Therefore, option A is correct. The field is not generally zero, and its axial and equatorial magnitudes are also different.

The first step is to classify the surface as open or closed and then identify the enclosed charge. For an open surface, flux is calculated from the field and area-vector relation. For a closed surface, Gauss’s law gives total flux as Q_enclosed divided by epsilon-zero. In a dipole problem, enclosing both charges gives zero net charge, while enclosing only one gives nonzero flux. Hence option A is correct.

A complete electric dipole contains equal positive and negative charges, so its net charge is zero. Gauss’s law therefore gives zero net flux through any closed surface enclosing the whole dipole. However, the two charges are separated spatially, and their individual electric fields generally do not cancel at every point on the surface. Thus the local field can be nonzero while the signed total flux is zero, making A correct.

In a uniform field, the positive and negative charges of a dipole experience forces of equal magnitude and opposite direction, so their vector sum is zero, although a torque may act. In a non-uniform field, the field strength differs at the two charge positions. Since F = qE, the two forces can have different magnitudes, producing a nonzero resultant force.

A dipole contains equal and opposite charges separated by a finite distance, so its net charge is zero but its forces need not cancel in a non-uniform field. The field values at the positive and negative ends can differ, and F = qE gives unequal force magnitudes. Their vector sum can therefore pull the dipole toward the region of stronger field; neutrality does not imply zero force.

For a short dipole at a far axial point, the field magnitude is approximately E_axial = (1/4πε₀)(2p/r³), so it varies as 1/r³. If the distance changes from r to 2r, the new field is proportional to 1/(2r)³ = 1/(8r³). Hence E_new/E_old = 1/8. The inverse-square law would give one fourth, but that applies to a point charge, not a far dipole field.

At a far equatorial point, the dipole field magnitude is E_equatorial = (1/4πε₀)(p/r³), apart from its direction being opposite to the dipole moment. Thus its magnitude varies as 1/r³. Replacing r by 3r gives E_new/E_old = r³/(3r)³ = 1/27. Therefore one twenty-seventh remains. One ninth would incorrectly use an inverse-square dependence.

For a short dipole at a far point, the axial field magnitude is E_a = (1/4πε₀)(2p/r³), whereas the equatorial field magnitude is E_e = (1/4πε₀)(p/r³). Dividing gives E_a/E_e = 2. The directions are different—the equatorial field is opposite to the dipole moment—but the question asks for magnitudes. Therefore the axial magnitude is twice the equatorial magnitude.

Consider a point on the perpendicular bisector of the dipole. It is equally distant from the positive and negative charges, so the components perpendicular to the dipole axis cancel by symmetry. The components along the axis point toward the negative charge and add. Since the dipole moment is defined from negative to positive charge, the resultant equatorial field is opposite to p. Hence option A is correct.

A point charge has a nonzero net charge, so its far electric field varies as 1/r². A dipole contains equal and opposite charges, whose leading 1/r² contributions cancel at distances much larger than the separation. The remaining dipole term varies approximately as p/r³. Therefore the dipole field decreases faster with distance. The cancellation, not the absence of charge, is the governing reason.

For a point on the far axial line of a dipole, the field is proportional to p/r³, specifically E ≈ 2kp/r³. If p becomes 2p and r becomes 2r, the ratio is E′/E = 2/(2³) = 2/8 = 1/4. Thus the new field is one-fourth of the original field. The distance effect is cubic and outweighs the doubling of dipole moment.

The far electric field of a dipole varies as E ∝ p/r³. On changing p to 3p and r to 3r, the field ratio becomes E′/E = 3/(3³) = 3/27 = 1/9. Therefore one-ninth of the original far field remains. One-third would incorrectly use an inverse-first-power distance law, while three times ignores the cubic decrease with distance.

A point charge produces an electric field proportional to 1/r². A dipole contains equal and opposite charges separated by a small distance. At a far point, the leading 1/r² contributions from the two charges almost cancel because their charges are opposite. The remaining dipole contribution is proportional to p/r³, so it decreases faster with distance. Therefore A is correct; both charges create fields, and the point-charge field is not distance-independent.

Take a point on the perpendicular bisector of the dipole. It is equally distant from +q and −q, so the two field magnitudes are equal. Their components perpendicular to the dipole axis cancel by symmetry, while the components along the axis point in the same direction, from the positive side toward the negative side. Since the dipole moment is defined from negative to positive charge, the resultant field is opposite to p. Hence A is correct.

The dipole moment p is defined from the negative charge to the positive charge. For a point on the axial line beyond the positive charge, the field due to +q points away from +q, while the field due to −q points toward −q; both contributions point along the negative-to-positive direction at that external point. Their resultant therefore has the direction of p. The far-point condition permits the usual dipole approximation; the field is not zero and is not equal to the equatorial field.

Let the far-field magnitude be E ∝ 1/rⁿ. If distance changes from r to 2r, the new field is E/2ⁿ. The problem states that it becomes eight times smaller, so 2ⁿ = 8 = 2³ and n = 3. Therefore E ∝ 1/r³, which is the characteristic far-field dependence of an electric dipole. Option A is correct; option B describes a point-charge field, while C and D show incorrect dependence.

For a point charge, the electric field decreases as E_point ∝ 1/r². For a dipole at distances much larger than its separation, the leading field decreases as E_dipole ∝ 1/r³. Since 1/r² falls more slowly than 1/r³ as r increases, the point-charge field remains relatively stronger at large distances. Neither field becomes instantly zero, so A is correct.

In the far-field approximation, the axial dipole field has magnitude E_axial = 2kp/r³, while the equatorial field has magnitude E_equatorial = kp/r³, with appropriate directions. Their magnitudes therefore differ by a factor of two, but both contain the same inverse-cube dependence on distance. They are not equal, parallel in direction, or always zero, so option A is correct.