Electric Flux and Electric Dipole

Flux is the dot product of electric field and the directed area element: dΦ = E·dA. The area vector of a closed surface is conventionally outward. When the electric field leaves the surface, it points in the same direction as the outward area vector, so the angle between them is 0° and cos 0° = 1. Therefore the flux contribution is positive, not negative or automatically zero.

Electric flux through a plane surface is Φ = EA cos θ, where θ is the angle between the electric field vector E and the area vector A, which is normal to the surface. Thus the relevant angle is not directly the angle between E and the surface itself. If the angle with the surface is supplied, it must be converted appropriately before using the cosine relation.

For a uniform electric field crossing a plane surface, flux is Φ = EA cos θ. If the electric field E and orientation θ remain unchanged, then cos θ is constant and flux is directly proportional to area A. Replacing A by 2A gives Φ′ = E(2A)cos θ = 2Φ. Therefore the flux doubles. It would not remain unchanged or become zero unless another quantity changed.

For a fixed surface and fixed orientation, the electric flux is Φ = EA cos θ. Here the area A and angle θ do not change, so the factor A cos θ remains constant. If the field changes from E to 2E, the new flux is Φ′ = (2E)A cos θ = 2Φ. Hence flux doubles. It becomes four times only if both the field and another independent proportional factor also double.

For a plane open surface in a uniform field, flux is Φ = EA cos θ, where θ is measured between the field and the area vector. The area vector is perpendicular 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. Therefore no field passes normally through the surface and the flux is zero.

An electric dipole consists of two equal and opposite charges separated by distance d. Its dipole moment is defined as p = qd in magnitude, where q is the magnitude of either charge and d is the separation between the charges. The vector points from the negative charge to the positive charge. Thus multiplying charge magnitude by separation is correct; division or either quantity alone is incomplete.

The magnitude of the dipole moment is p = qd, where q is the magnitude of either charge and d is the separation. In this question, d remains constant while q changes to 2q. Therefore the new moment is p′ = (2q)d = 2qd = 2p. The dipole moment consequently doubles. It does not become zero or unchanged, because the charge magnitude has changed directly.

The electric dipole moment is defined by p = qd, where q is the magnitude of either charge and d is the separation between the charges. Since q remains unchanged while d becomes d/2, the new moment is p′ = q(d/2) = p/2. Therefore, option A is correct. It does not become double or four times, and it becomes zero only if the separation becomes zero.

A polar molecule has an unequal distribution of charge, so the centres of its positive and negative charges are separated. This separation gives the molecule a permanent electric dipole moment, provided the molecule is considered in its stable structure. Hence option A is correct. A polar molecule is not merely a single positive or negative charge, and it does not have zero charge separation like the distractors suggest.

The electric field of a finite dipole decreases as the observation point moves farther away. In the far-field region, its magnitude varies approximately as E ∝ 1/r³, unlike the 1/r² dependence of an isolated point charge. The fields of the two opposite charges partly cancel at large distance, so option A is correct. It neither remains constant nor becomes infinite.

An electric dipole consists of two point charges having equal magnitudes and opposite signs, written as +q and −q, separated by a small distance. Thus the magnitudes are both q, even though the algebraic signs differ. Option A is correct. The charges are not necessarily double or zero, and equal same-sign charges would form a different charge arrangement rather than an ideal dipole.

For a uniform field over a plane surface, electric flux is Φ = EA cos θ, where θ is the angle between the electric field and the area vector, which is normal to the surface. If the field is parallel to the surface, θ = 90°, so cos θ = 0 and Φ = 0. Therefore option A is correct. A field along the area vector gives maximum magnitude, not zero flux.

Electric flux through a plane surface is given by Φ = EA cos θ, where θ is measured between the electric field and the area vector. For θ = 90°, cos 90° = 0, so Φ = EA × 0 = 0. Hence option A is correct. Maximum positive flux occurs at 0°, maximum negative flux at 180°, while neither condition produces infinite flux in this context.

The flux relation is Φ = EA cos θ. At θ = 180°, the electric field is opposite to the chosen outward area vector, and cos 180° = −1. Therefore Φ = −EA, so the flux is negative; option A is correct. It is not zero, because zero occurs at 90°. The sign depends on the chosen area-vector orientation, while the stated orientation fixes it as negative.

Electric dipole moment is defined as p = qd, where its direction is conventionally taken from the negative charge to the positive charge. Thus it has both magnitude and a definite direction, making it a vector quantity. Option A is correct. It is not scalar or directionless, and it is not unitless: its SI unit is coulomb-metre (C m).

Electric flux through a plane surface in a uniform electric field is Φ = EA cos θ, where E is the field magnitude, A is the surface area, and θ is the angle between the field and the outward area vector. Thus option A gives the correct dependence. Flux is maximum when θ = 0° and zero when θ = 90°. The other options have incorrect dimensions or omit orientation.

For a plane surface in a uniform field, electric flux is Φ = EA cos θ. If the area A and orientation θ remain unchanged, flux is directly proportional to the field magnitude E. Therefore, increasing E increases Φ in the same ratio. Option A is correct. Flux would decrease only if E decreased or if the orientation changed appropriately; it does not automatically become zero or remain unchanged.

The magnitude of an ideal electric dipole moment is p = qd, where q is the magnitude of either charge and d is the separation vector magnitude. If p = 0, the ideal dipole effect is absent; this can occur when the charge or separation is zero, or when no effective dipole exists. Therefore option A is the appropriate general statement. The other choices incorrectly claim that zero moment means large charge or separation.

The dipole moment magnitude is p = qd, the product of charge magnitude and separation. Initially p = qd. After doubling both quantities, p′ = (2q)(2d) = 4qd = 4p. Therefore the dipole moment becomes four times its original value, so option A is correct. Doubling only one factor would make it double, but both independent factors are doubled here.

Electric flux through a surface is Φ = EA cos θ, where θ is measured between the electric field and the chosen area vector. When the field and area vector point in the same direction, θ = 0° and cos θ = 1, giving positive flux. Thus option A is correct. Opposite directions give negative flux, while a field parallel to the surface gives zero flux because θ = 90°.

An electric dipole consists of equal and opposite charges separated by a small distance. The straight line passing through, and joining, the two charges is called the axial line or dipole axis. Therefore option A is correct. The perpendicular bisector through the midpoint is instead called the equatorial line, so option B describes the other standard dipole line.

The axial line of a dipole is the line joining its positive and negative charges. The equatorial line is defined geometrically as the straight line through the midpoint of the dipole and perpendicular to its axial line. Every point on this line is equidistant from the two charges. Hence option A is correct, while option B identifies the axial line.

The electric dipole moment points from the negative charge to the positive charge. On the axial line, the fields due to the two charges act along the same general axial direction at a distant point, and their resultant has the same direction as the dipole moment. Thus option A is correct. The field is not always zero, and the equatorial line is a different location where the direction reverses.

For a dipole, the dipole moment is directed from the negative charge to the positive charge. At a point on the equatorial line, the components of the two charge fields perpendicular to the dipole axis cancel, while the axial components add in the direction opposite to the dipole moment. Therefore option A is correct. The field is generally non-zero, so option D is incorrect.

Electric flux describes the amount of electric field passing through a specified surface. For a uniform field and a plane surface, it is expressed as Φ = EA cos θ, where E is field strength, A is area, and θ is the angle between the field and the area vector. Hence option A gives the correct basic meaning. Current, temperature, and mass are different physical quantities and cannot define electric flux.