Electric Flux and Electric Dipole

Electric flux is the dot product of electric field and area vector: Φ = E · A = EA cos θ. A dot product produces a scalar, so flux has magnitude and may have a positive or negative sign, but it does not possess an independent vector direction. Therefore option B is correct. Option A confuses flux with the electric field or area vector, while C and D do not describe a valid classification.

Electric flux is Φ = E A cos θ. The SI unit of electric field is N/C and the unit of area is m²; therefore the unit of flux is (N/C)m² = N m²/C. Option A is correct. Option B is only the unit of electric field, option C is its reciprocal, and option D is the unit of electric potential, not electric flux.

For a uniform electric field through a plane surface, flux is Φ = EA cos θ, where θ is measured between the field and the area vector, which is normal to the surface. The flux is maximum when θ = 0°, so the field is parallel to the area vector and therefore perpendicular to the surface. Thus option A is correct. A parallel field gives zero flux, while zero area or zero field also gives zero.

The flux through a plane surface is Φ = EA cos θ, with θ measured between the electric 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 90° = 0. Therefore option A is correct. A perpendicular field generally gives maximum flux, and a large area does not make flux zero; a closed surface alone does not guarantee zero net flux.

For a uniform field crossing a plane surface, electric flux is Φ = EA cos θ. Consequently, it depends on the electric-field strength E, the surface area A, and the orientation represented by θ, the angle between the field and area vector. Option A is correct. Temperature, mass, and colour are not direct variables in this basic flux expression, so options B, C, and D are irrelevant distractors.

An area vector represents both the magnitude of area and the orientation of a surface. For a plane surface, its direction is defined along the normal, meaning perpendicular to the surface. This direction is needed in Φ = E · A to determine the angle and sign of flux. Therefore option A is correct. The vector is not parallel to the surface and has no universally fixed eastward or downward direction.

For every small element of a closed surface, the area vector is perpendicular to that element. By the standard convention used in Gauss’s law, it points outward from the volume enclosed by the surface. This consistent choice makes the sign of outward and inward electric flux meaningful. Hence option A is correct. It is not always inward, cannot be parallel to the surface, and is not chosen randomly.

An electric dipole is a system of two point charges having equal magnitudes, opposite signs, and a small separation. It is commonly written as +q and −q separated by distance d, with dipole moment p = qd directed from the negative charge to the positive charge. Two like charges do not form a dipole, and a single charge is not a dipole. Therefore option A is correct.

An ideal electric dipole consists of charges +q and −q. Adding their algebraic values gives Q_net = (+q) + (−q) = 0, so its net charge is zero. However, zero net charge does not mean that the dipole has no electric field; the separated charges create a nonzero dipole moment p = qd and can influence nearby charges. Hence option A is correct.

Electric dipole moment is a vector defined by p = qd, where d is the displacement vector directed from the negative charge to the positive charge. Therefore its conventional direction is from −q to +q. It is not automatically upward, and it is not always opposite to the external electric field; its relation to the field depends on the dipole’s orientation. Thus option A is correct.

For a dipole made of charges +q and −q separated by distance d, the magnitude of dipole moment is p = qd. Thus it depends on both the magnitude of either charge and the separation between the charges. Its unit is coulomb metre. It is not merely charge or distance, and mass is irrelevant to this electrostatic quantity. Therefore option A is correct.

The magnitude of electric dipole moment is p = qd. Charge q is measured in coulombs and separation d in metres, so the SI unit is coulomb metre, written C m. Newton per coulomb is the unit of electric field, coulomb per metre is a linear charge-density unit, and newton metre represents torque or energy. Hence option A is the only correct unit.

The magnitude of an electric dipole moment is defined by p = qd, where q is the magnitude of either charge and d is the separation between the positive and negative charges. Thus increasing q or increasing d increases p in direct proportion. Option A is correct. Colour, area, and mass do not enter this electrostatic definition, so the remaining options cannot increase the dipole moment by themselves.

The electric dipole moment is p = qd, where q is the charge magnitude and d is the separation. Initially p = qd. If the charge stays unchanged and the separation becomes 2d, the new moment is p′ = q(2d) = 2qd = 2p. Therefore option A is correct. It is not half or zero, and it becomes four times only if both q and d are doubled.

For an electric dipole, p = qd, where q is the magnitude of either charge and d is the fixed separation. Initially p = qd. After each charge magnitude is reduced to q/2, with d unchanged, the new dipole moment is p′ = (q/2)d = p/2. Hence option A is correct. Doubling or quadrupling would require an increase in charge or another parameter, while unchanged p is inconsistent with direct proportionality.

The axial line of an electric dipole is the straight line passing through both unlike charges. The dipole moment is directed from the negative charge toward the positive charge along this same line. Therefore, option A is correct. Option B describes the equatorial line only when it passes through the midpoint and is perpendicular to the axis; arbitrary circles or field-parallel lines do not define an axial line.

The equatorial line is defined geometrically as the line through the midpoint of the dipole and perpendicular to its axial line. Every point on this line is equidistant from the positive and negative charges, which is useful in analysing the dipole field. Hence option A is correct. Option B is the axial line, while arbitrary lines from either charge have no such definition.

For a short electric dipole, the far-field axial expression is E_axial = (1/4πε₀)(2p/r³), directed along the dipole moment on the positive axial side. Thus the axial electric field is parallel to p, so option A is correct in the stated general sense. It is not always perpendicular, arbitrary, or zero; its magnitude decreases as 1/r³ but does not vanish merely because the point is far.

Consider a point on the equatorial line, where distances from +q and −q are equal. The two field contributions have equal magnitudes; their transverse components cancel, while their components along the dipole axis add from the positive charge toward the negative charge. Since p points from negative to positive charge, the resultant field is opposite to p. Therefore option A is correct.

Electric flux is defined by Φ = ∫ E·dA, where dA is the chosen area vector. Positive flux means the net field component is in the same direction as that area vector; in the uniform case, Φ = EA cosθ is positive when cosθ > 0. Thus field lines cross in the selected positive sense, making option A correct. Positive flux does not mean zero field or negative charge.

Flux is the surface integral Φ = ∫ E·dA, so its sign comes from the dot product of the field with the selected area vector. Negative flux occurs when the net field component points opposite to that vector; for a uniform field, this means cosθ < 0. Therefore option A is correct. Negative flux does not imply no field, zero area, or necessarily positive charge.

For a uniform electric field, flux through a planar surface is Φ = EA cosθ. In a favourable orientation, the field and area vector are aligned, so θ = 0 and Φ = EA. With E and orientation fixed, flux is directly proportional to area A; increasing A therefore increases Φ. Option A is correct. It would not decrease or become zero, and it remains unchanged only if the area is unchanged.

The electric flux through a surface is Φ = EA cosθ for a uniform field. Since the area A and orientation angle θ remain unchanged, the factor A cosθ is constant. If E changes to 2E, the new flux is Φ′ = (2E)A cosθ = 2Φ. Hence option A is correct. It is not halved or quadrupled because flux is first-order, directly proportional to field magnitude.

Electric flux is Φ = EA cosθ, where θ is the angle between the electric field and the area vector. Rotating the surface changes the direction of its area vector, so θ changes even though E and A retain the same magnitudes. Consequently, cosθ and the flux can change. Option A is correct; rotation does not alter charge units, mass, or turn abstract field lines into wires.

The electric field of a dipole becomes weaker as the observation point moves farther away. In the far-field region, the dipole field varies approximately as 1/r³, where r is the distance from the dipole. Thus increasing r reduces the field magnitude, so option A is correct. It is not generally constant or always zero; only at selected symmetry points can cancellation occur.