Electric Flux and Electric Dipole

The correct answer is A. A permanent electric dipole moment is defined by p = qd, where q represents the effective separated charge and d is the displacement from the negative-charge centre to the positive-charge centre. In a polar molecule, unequal sharing of electrons and molecular structure keep these charge centres separated. Colour, mass, or the presence of only positive charge does not create a permanent dipole moment.

The correct answer is A. The permanent dipole moment is p = qd, with d measured between the effective centres of positive and negative charge. In an ideal non-polar molecule, the charge distribution is symmetric, so these centres coincide and d = 0; consequently p = 0. Such a molecule may develop an induced dipole in an external field, but that temporary effect does not give it a permanent dipole moment.

The correct answer is A. For a dipole of moment p, at a distant axial point the field magnitude is approximately E_axial = (1/4πε₀)(2p/r³), while at a distant equatorial point it is E_equatorial = (1/4πε₀)(p/r³). At the same large distance r, their ratio is therefore E_axial/E_equatorial = 2. The directions differ, but the question asks for magnitude comparison.

The correct answer is A. Consider a point on the perpendicular bisector of a dipole. The point is equally distant from +q and −q, so the field components perpendicular to the dipole axis cancel by symmetry. The components along the dipole axis add in the direction of the negative charge. Since the dipole moment is defined from negative to positive charge, the resultant equatorial field is opposite to p, with magnitude approximately p/(4πε₀r³) far away.

The correct answer is A for a far axial point on the side of the positive charge. The dipole moment p points from the negative charge to the positive charge. On the axial line beyond the positive charge, the field due to the nearer positive charge and the field due to the farther negative charge combine in the direction of p. The far-field expression E_axial ≈ (1/4πε₀)(2p/r³) confirms this direction.

An electric dipole consists of equal and opposite charges separated by a small distance. At a far point, the fields due to these charges are nearly equal and opposite, so their leading contributions cancel. A point charge has a field proportional to 1/r², whereas the far dipole field is proportional to 1/r³. Therefore, option A correctly explains the faster decrease; the other statements contradict the definition of charge and electric field.

The dipole moment of an ideal dipole is p = qd, where q is the magnitude of either charge and d is the separation vector. If p = 0, then either the charge magnitude is zero, the effective separation is zero, or separate contributions cancel in a more general system. Thus option A is the valid conclusion. A zero moment does not imply a strong or infinite dipole, and large separation would generally increase the moment.

Electric flux is the surface integral Φ = ∫ E·dA. Its magnitude describes the net amount of electric field passing through the surface, while its sign is determined by the chosen area-vector direction: outward for a closed surface or a specified normal for an open one. Positive and negative flux therefore indicate opposite directional senses, not simply more and fewer field lines. Option A gives the correct interpretation.

Gauss’s law states that the net flux through a closed surface is Q_enclosed divided by epsilon-naught. A dipole has equal positive and negative charges, so Q_enclosed = 0 and the algebraic sum of flux is zero. However, the two separated charges produce nonzero fields at most exterior points; flux entering and leaving can cancel without the local field cancelling. Thus option A is correct; the other choices misuse Gauss’s law or describe field lines incorrectly.

Flux is the dot product Phi = E dot A = EA cos theta. The area vector supplies the chosen normal direction of the surface. When the electric field has a component along that chosen outward direction, the flux is positive; when it points opposite to it, the flux is negative. Therefore option A is correct. Surface colour, charge mass, and thickness do not determine the sign of electrostatic flux.

Gauss’s law states that the net electric flux through a closed surface is Φ = Q_enclosed/ε₀. An electric dipole contains equal charges +q and −q, so its net enclosed charge is q − q = 0. Therefore the total flux is zero, even though the electric field is nonzero at many points on the surface. Positive or negative flux would require a nonzero net enclosed charge, and infinite flux is not applicable here.

By Gauss’s law, the net flux through a closed surface is determined only by the algebraic charge enclosed: Φ = Q_enclosed/ε₀. If the surface contains only +q, then Q_enclosed = +q and Φ = +q/ε₀, so the flux is positive. The negative partner being outside does not cancel this flux. The dipole moment is not the deciding quantity for net closed-surface flux.

Gauss’s law gives the net electric flux through a closed surface as Φ = Q_enclosed/ε₀. When only the negative charge −q of the dipole lies inside, Q_enclosed = −q, so Φ = −q/ε₀. Thus the outward-oriented net flux is negative. It is not zero because the opposite charge is outside, and it is not necessarily maximum; the enclosed charge alone fixes the net value.

For an ideal electric dipole consisting of +q and −q separated by distance d, the magnitude of dipole moment is p = qd. Thus it depends on the magnitude of either charge and the separation between the charges. The net charge is zero, so the sum of charges alone cannot specify p. An external electric field and the area of an unrelated surface are not required to define the dipole moment.

The dipole-moment magnitude is p = qd, where q is the magnitude of either charge and d is the separation. Initially p = qd. After the changes, q′ = 2q and d′ = d/2, so p′ = q′d′ = (2q)(d/2) = qd = p. The doubling and halving exactly cancel; therefore the dipole moment remains unchanged.

For an electric dipole, p = qd. Let the original moment be p = qd. With the stated changes, q′ = q/2 and d′ = 2d. Therefore p′ = q′d′ = (q/2)(2d) = qd = p. The reduction in charge is exactly compensated by the increase in separation, so the dipole moment remains unchanged rather than becoming double, half, or four times.

Electric dipole moment is a vector defined as p = qd, where the displacement vector d is taken from the negative charge to the positive charge. Hence its direction is from −q to +q. Electric field lines conventionally run from positive to negative, so option B reverses the definition. The dipole-moment direction is not always opposite to an external field and has no general connection with an area vector.

For a dipole with moment p directed from −q to +q, an axial point far from the dipole lies on the same line as p. The axial field magnitude is approximately E = (1/4πε₀)(2p/r³), and its vector direction is along p on the positive axial side. Thus the far axial field is along the dipole moment. The opposite direction applies to the equatorial field, not the axial field.

An equatorial point lies on the perpendicular bisector of the dipole. The fields from +q and −q have equal components along the equatorial direction that cancel, while their components along the direction opposite to p add. Consequently, the far equatorial field has magnitude approximately E = (1/4πε₀)(p/r³) and points opposite to the dipole moment. It is nonzero, so option D is incorrect.

For a short electric dipole, the far axial field is E_axial = (1/4πε₀)(2p/r³), whereas the equatorial field has magnitude E_equatorial = (1/4πε₀)(p/r³). At the same distance r, their ratio is therefore 2:1. Hence the axial field is double the equatorial field. The options half, equal and four times do not match this ratio.

Each charge experiences force F = qE at its own position. In a non-uniform field, the field values at the positive and negative charges can differ because the charges are separated in space. Consequently, the magnitudes of the two opposite-directed forces need not be equal, leaving a nonzero resultant force. This differs from a uniform field, where the forces cancel exactly.

The net charge of a dipole is q + (−q) = 0, but its dipole moment is a vector defined by p = qd, where d is the separation vector from the negative to the positive charge. Since the equal opposite charges occupy different positions and d is nonzero, p is nonzero. If the charges were at the same point, only then would the dipole moment vanish.

The governing idea is the distance dependence of electric fields. For a point charge, Coulomb’s law gives E = kQ/r², so the field decreases with the square of distance. A dipole has zero net charge, and its leading far-field term is proportional to p/r³, where p is the dipole moment. Therefore option A is correct; options B, C, and D reverse or misstate these dependences.

Electric flux through a plane surface is Φ = EA cos θ, where θ is measured from the chosen area vector to the electric field. A negative value means cos θ < 0, so the field has a component opposite to that vector, usually with an angle greater than 90°. It does not mean that the field or area is zero or that the source charge must be negative. Thus option A is correct.

The equatorial line of a dipole is defined geometrically as the straight line passing through the midpoint between the two charges and perpendicular to the line joining them. The line joining the charges is the axial line, so option B describes the axial direction rather than the equatorial one. Options C and D describe possible electric-field-line behavior, not the definition of the dipole’s equatorial line. Therefore, option A is correct.