Electric Flux and Electric Dipole

Electric flux is Φ = EA cos θ, while maximum positive flux is Φmax = EA when θ = 0°. For θ = 30°, Φ = EA cos 30° = (√3/2)EA, approximately 0.866 EA. This is positive but smaller than EA, so option A is correct. It is not zero because the vectors are not perpendicular, cannot exceed EA for fixed E and A, and is not negative because cos 30° is positive.

The sign of electric flux follows Φ = EA cos θ. It is positive when the angle θ between the electric field and the chosen area vector is less than 90°, zero at 90°, and negative when θ exceeds 90° because cosine then becomes negative. Thus changing the orientation can change flux from positive to negative, making option A correct. Increasing E or A changes magnitude, not sign by itself, and a plane surface does not determine the sign.

A permanent dipole moment requires a fixed separation between the centres of positive and negative charge. In a non-polar molecule, symmetry generally makes these charge centres coincide, so the separation vector and permanent dipole moment are zero. Such a molecule may acquire an induced dipole in an external field, but that temporary effect does not create a permanent dipole. Therefore option A is correct.

Gauss’s law states that the total electric flux through any closed surface is Φ = Q_enclosed/ε₀. A complete dipole contains equal charges +q and −q, so its net enclosed charge is Q_enclosed = q − q = 0. Consequently Φ = 0, regardless of the dipole’s position or orientation inside the surface. The flux is not equal to the dipole moment, which is a vector quantity.

For an ideal electric dipole, the magnitude of dipole moment is p = qd, where q is the magnitude of either charge and d is the separation between the charges. Its direction is defined from the negative charge to the positive charge. Since a physical quantity having both magnitude and a specified direction is a vector, electric dipole moment is treated as a vector. It is not always zero and has unit C m.

The magnitude of an electric dipole moment is p = qd, where q is the magnitude of either charge and d is the separation between the charges. Initially p = qd. After doubling the separation, d′ = 2d, while q remains unchanged; therefore p′ = q(2d) = 2qd = 2p. The dipole moment consequently becomes twice its original value. Option A is correct. It is not unchanged because separation enters directly, and it is not four times because only one factor, d, was doubled.

For an electric dipole, the moment magnitude is p = qd. Let the original values be q and d, so p = qd. The changes give q′ = 3q and d′ = d/2. Therefore p′ = q′d′ = (3q)(d/2) = 3qd/2 = 1.5p. Thus the new dipole moment is one and a half times the original, making option A correct. Multiplying only the charge factor would suggest three times, while multiplying both changes incorrectly as six ignores that the separation is reduced.

The electric field-line direction and the dipole-moment direction follow different conventions. Field lines outside a dipole point from the positive charge toward the negative charge. In contrast, the dipole moment vector p is defined from the negative charge to the positive charge. Thus option A is correct. Option B describes the usual external field-line direction, not p; options C and D are unrelated because no surface is specified and a dipole moment is generally nonzero.

For a uniform electric field crossing a plane surface, electric flux is Φ = EA cos θ, where E is the field magnitude, A is the area, and θ is the angle between the field and the area vector. Since E and θ remain unchanged, Φ is directly proportional to A. Replacing A by 3A gives Φ′ = E(3A)cos θ = 3Φ. Therefore, option A is correct; the other ratios do not follow the flux formula.

Electric flux through a plane surface is Φ = EA cos θ. With the orientation unchanged, cos θ is constant, so the flux changes according to the product EA. The new field is E/2 and the new area is 2A; hence Φ′ = (E/2)(2A)cos θ = EA cos θ = Φ. The two scale factors cancel exactly. Therefore option A is correct, while the other options incorrectly apply only one factor or multiply them in the wrong way.

The governing distinction is between three different ideas: net flux through a closed surface depends on enclosed charge, the local field describes the vector field at a point, and dipole moment points from negative charge to positive charge. A zero net flux does not imply zero field. Therefore mixing these ideas is the common mistake, so A is correct. The other options are useful steps or routine checks, not mistakes.

For a uniform field crossing a plane surface, electric flux is Phi = E A cos theta, where theta is the angle between the field and the area vector, which is normal to the surface. Flux is maximum when cos theta = 1, or theta = 0 degrees. Therefore the field is perpendicular to the surface and parallel to the area vector. Hence A is correct; a parallel field gives zero flux.

Electric flux through a plane is Phi = E A cos theta, with theta 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 theta = 90 degrees and cos theta = 0. Thus Phi = 0. Option A is correct. A perpendicular field generally gives maximum flux, and merely being closed does not by itself make flux zero.

For a uniform electric field, the flux through a plane surface is Phi = E A cos theta. If the field magnitude E and the orientation angle theta remain fixed, then cos theta is constant and Phi is directly proportional to area A. Replacing A by 2A gives Phi' = E(2A)cos theta = 2Phi. Therefore option A is correct; no inverse or square dependence is involved.

Electric flux is given by Phi = E A cos theta. In this question, the area A and the angle theta are fixed, so A cos theta is constant. If the field changes from E to 2E, the new flux is Phi' = (2E)A cos theta = 2Phi. Thus the flux doubles, making option A correct. The unchanged, half, and zero results would require a different change in the given quantities.

At a point far from a dipole, its electric-field magnitude follows the inverse-cube law, E ∝ 1/r³. If the distance changes from r to 2r, the new field is E' = E/(2³) = E/8. Thus the field becomes one-eighth of its former value, or it decreases by a factor of eight. Inverse-square and unchanged behaviors are not applicable here.

The electric field of a dipole at a far point obeys E ∝ 1/r³. On tripling the distance, r becomes 3r, so E' = E/(3³) = E/27. Therefore the field becomes one twenty-seventh of its initial value. The one-ninth and one-third choices correspond to different powers of distance, while tripling is opposite to the actual decrease.

An area vector represents both the magnitude of area and the orientation of a surface. For a plane surface, its direction is chosen perpendicular, or normal, to the plane; for a closed surface, the outward normal is conventionally used. It is not generally parallel to the surface, automatically aligned with the electric field, or always downward. Therefore option A is correct.

For a uniform field crossing a plane surface, flux is Φ = EA cos θ, with θ measured between the electric field and the area vector. Alignment means θ = 0°, so cos 0° = 1 and Φ = EA, its largest positive value for fixed E and A. A right angle would give zero flux, while opposite alignment would give maximum negative flux. Hence option A is correct.

Electric flux through a plane surface is Φ = EA cos θ, where θ is the angle between the electric field and the area vector. If the area vector is perpendicular to the field, θ = 90° and cos 90° = 0. Therefore Φ = 0. This does not mean the electric field itself is absent; its component along the area vector is zero. Thus option A is correct, while B and D correspond to parallel and antiparallel orientations.

An electric dipole is formed by two equal and opposite charges separated by a small distance. The axial line is the straight line passing through, and joining, the two charges; it is the direction of the dipole moment from −q to +q. The perpendicular bisector is instead called the equatorial line. Therefore option A is correct, while B describes the equatorial geometry and C and D are unrelated.

Electric flux through an open surface is defined by Φ = ∫ E · dA, or Φ = EA cos θ for a uniform field over a plane surface. Reversing the area vector changes dA to −dA, so the flux changes from Φ to −Φ while its magnitude remains unchanged. Since the original flux is negative, the new flux is positive. Therefore option A is correct; reversing orientation neither makes the flux zero nor doubles its magnitude.

For a closed surface, the area element dA is conventionally directed outward. In the flux integral Φ = ∮ E · dA, a field pointing outward has a positive dot product, whereas a field entering the surface has a negative dot product. This gives a consistent meaning to net flux and allows Gauss’s law, Φ = Q_enclosed/ε₀, to be applied without ambiguity. Thus option A is correct; the convention creates neither fields nor charges and does not require a spherical surface.

For a uniform field through a plane surface, electric flux is Φ = EA cos θ. At θ = 90°, cos θ = 0, so the flux is zero. When θ increases beyond 90°, cos θ becomes negative, meaning the field component along the chosen area-vector direction reverses sign. Consequently, the flux can change from positive to negative. Option A is correct; maximum positive flux occurs at 0°, and neither the unit nor the physical area changes merely because the angle changes.

Electric flux through a plane surface is Φ = E A cos θ, where θ is the angle between the electric field and the chosen area vector. Here the two vectors point in opposite directions, so θ = 180° and cos 180° = −1. Hence Φ = −EA, which is negative. It would be zero only if the field were parallel to the surface, not opposite to its area vector.