Electric Flux and Electric Dipole

For a dipole, the dipole moment points from the negative charge to the positive charge. At a point on the perpendicular bisector, the fields from the two charges have components perpendicular to the dipole axis that cancel, while their components along the axis point from positive to negative and add. Hence the equatorial field is opposite to the dipole moment, so option A is correct.

Flux is Φ = EA cos θ, where θ is measured between the electric field and the area vector, not directly between the field and the surface. When the surface is parallel to the field, its area vector is perpendicular to the field, so θ = 90° and flux is zero. When the surface is perpendicular, the area vector is parallel, θ = 0°, and flux is maximum. Thus A is correct.

An electric dipole consists of equal and opposite charges separated by a finite distance. Although their algebraic sum is zero, the electric fields produced by the two charges do not cancel at every point because their directions and distances from the observation point differ. Their vector sum is therefore nonzero at many locations. Option A correctly identifies spatial separation; the other choices make false claims about zero net charge or the field of one sign of charge.

The governing idea is the force relation F = qE. A dipole contains equal and opposite charges, but in a non-uniform field the two charges occupy different positions and generally experience different field magnitudes. Their forces are therefore not necessarily equal, so they cannot always cancel and a net translational force may remain. The dipole may also experience torque. Option A gives the correct reason; the dipole remains electrically neutral, negative charges do feel force, and flux can exist in a non-uniform field.

For a dipole, the dipole moment p is defined from the negative charge toward the positive charge. On the axial line, the fields produced by the two charges are collinear. At a point far from the dipole, their resultant has the direction of p, although its magnitude is much smaller than the field of a single charge and varies approximately as 1/r^3. Therefore option A is correct. The equatorial direction is not axial, and the field is not generally zero at a finite far point.

The governing principle is vector addition of the fields of the two equal and opposite charges. At an equatorial point, the components perpendicular to the dipole axis cancel by symmetry, while the components along the axis add in the direction opposite to the dipole moment. Thus the resultant field points opposite to p and has magnitude approximately proportional to 1/r^3 at a far point. Option A is correct; it is not along p and is not generally zero.

The governing concept is cancellation in the far-field approximation. A point charge produces an electric field proportional to 1/r². A dipole has equal and opposite charges, so at a distant point their leading 1/r² contributions largely cancel. The remaining dipole field is proportional to p/r³, where p is the dipole moment, and therefore decreases faster with distance. Option A is correct; a dipole does contain charges, they have opposite signs, and distance strongly affects the field.

Gauss’s law states that the total electric flux through any closed surface is Φ = Qenclosed/ε0. Therefore it depends on the net charge enclosed, not on the surface’s shape, size, or total area. If deformation does not change the enclosed net charge, the total flux remains unchanged. Option A is correct. It is not necessarily zero unless the net enclosed charge is zero, and it does not automatically double or depend only on area.

Electric field lines are drawn in the direction of the force on a positive test charge, so outside a dipole they point from the positive charge toward the negative charge. The dipole moment is a separate vector defined by convention as p = qd, directed from the negative charge to the positive charge. This definition is not a contradiction. Therefore option A is correct; the field-line direction and dipole-moment direction represent different conventions.

For a uniform field crossing a plane surface, electric flux is Φ = EA cos θ, where θ is the angle between the field and the area vector, perpendicular to the surface. If the electric field is parallel to the surface, θ = 90°, so cos 90° = 0 and Φ = 0. If the field were along the area vector, flux would be maximum, not zero. Thus A is the valid reason.

Electric flux measures how much electric field passes through a surface. For a uniform field it is Φ = EA cos θ, where E is field magnitude, A is surface area, and θ is the angle between E and the area vector. Thus two surfaces in the same field can have different flux because their areas or orientations differ. Field magnitude alone is insufficient, so option A is correct.

A dipole consists of equal and opposite charges, and the field at any point is the vector sum of the fields due to both charges. On the axial line, the relevant components reinforce in the dipole-moment direction. On the equatorial line, symmetry makes the transverse components cancel while the remaining component points opposite to the dipole moment. The different component combinations therefore produce different directions.

Electric flux is defined by Φ = ∫ E · dA. The dot product becomes negative when the electric-field component along the chosen area vector points in the opposite direction. Thus, negative flux conveys directional information; it does not mean that the field, area, or surface is nonexistent. Therefore option A is correct, while B, C, and D confuse a sign convention with physical absence.

Gauss’s law states that the net electric flux through a closed surface equals the enclosed charge divided by ε₀. A complete electric dipole contains equal charges +q and −q, so its net enclosed charge is q − q = 0. Consequently, the total closed-surface flux is zero, although the electric field and local flux may be nonzero. Hence A is correct.

Electric flux through a plane surface is given by Φ = EA cos θ, where θ is the angle between the electric field and the area vector. Since the field is perpendicular to the surface, it is parallel to the area vector, so θ = 0° and cos 0° = 1. Therefore Φ = 5 × 2 × 1 = 10 N m²/C. Hence option A is correct; B omits the area, C divides instead of multiplying, and D would apply to a parallel field.

The governing relation is Φ = EA cos θ, with θ measured between the electric field and the area vector, which is normal to the surface. If the field is parallel to the surface, it is perpendicular to the area vector, so θ = 90° and cos 90° = 0. Thus Φ = 0. Option A is correct. Maximum positive or negative flux occurs when the field is parallel or antiparallel to the area vector, not parallel to the surface.

Electric flux depends on both the magnitude of the surface area and its orientation relative to the electric field. This is expressed by Φ = E · A = EA cos θ. The area vector has magnitude equal to the area and direction normal to the surface, allowing the angle θ and the sign of flux to be represented. Therefore option A is correct. Area is not inherently negative, electric field is a vector, and charge colour has no physical role.

Electric flux is Φ = EA cos θ. For θ = 0°, the electric field and area vector point in the same direction, and cos 0° = 1. Hence Φ = EA, its greatest possible positive value for fixed E and A. Option A is correct. Zero flux occurs at 90°, maximum negative flux occurs at 180°, and a half value would require cos θ = 1/2, not θ = 0°.

The sign follows from Φ = EA cos θ. At θ = 180°, the electric field and area vector point in opposite directions, and cos 180° = −1. Thus Φ = −EA, so the flux is maximum negative for the given field and area. Option A is correct. Positive flux corresponds to θ = 0°, while zero flux corresponds to θ = 90°; the sign is therefore fully determined here.

Gauss’s law states that the net electric flux through a closed surface is Φ = Q_enclosed/ε₀. The enclosed charge is Q_enclosed = +2 C − 2 C = 0 C. Therefore Φ = 0/ε₀ = 0. Option A is correct. The individual charges may produce electric fields and local flux through different parts of the surface, but their total enclosed charge cancels, so the net flux is neither positive nor negative.

A complete electric dipole consists of equal charges +q and −q, so its enclosed net charge is Q_enclosed = q − q = 0. Gauss’s law states that the net flux through a closed surface is Φ = Q_enclosed/ε₀; therefore Φ = 0. Option A is correct. The dipole does produce an electric field, and flux can enter and leave different portions of the surface, but the total signed flux cancels because the enclosed charge is zero.

Electric dipole moment is a vector defined by convention as p = qd, with its direction from the negative charge toward the positive charge. This direction is not the same as the direction of electric field lines, which generally go from positive to negative outside the dipole. Option A is therefore correct. The other statements are false: positive charges do produce electric fields, the two dipole charges have equal magnitudes, and a dipole moment definitely has a direction because it is a vector quantity.

Electric flux through a surface is Φ = E · A = EA cos θ, where A is the oriented area vector. For an open surface, the choice of normal direction is arbitrary. Reversing the area vector changes θ to 180° − θ and changes Φ to −Φ, without necessarily changing the physical field or surface area. Therefore option A is correct; area magnitude, naming, or colour cannot reverse the flux sign.

For a uniform field and a plane surface, electric flux is Φ = EA cos θ, where θ is the angle between the electric field and the area vector normal to the surface. Rotating the surface rotates this vector, so θ changes and the cosine factor changes. Consequently, the flux varies even when E and the area remain constant. Thus option A is correct; no charge mass, material change, or disappearance of field lines is required.

The correct answer is A. An electric dipole consists of equal and opposite charges separated by a finite distance. Although their algebraic sum is zero, the electric fields produced by the two charges do not cancel at every point because the charges occupy different positions. Their vector combination creates a nonzero field, especially near the dipole. Net charge determines the far leading term, but charge separation determines the dipole field.