Electric Flux and Electric Dipole

Electric flux through a plane surface is Φ = E · A = EA cos θ, where A is the area vector and θ is the angle between E and A. Reversing the area-vector direction changes A to −A, so the flux becomes Φ′ = E · (−A) = −Φ. Its magnitude remains the same, but its sign reverses. Therefore a positive flux becomes negative. It does not double, vanish, or remain unchanged, so option A is correct.

The dipole moment points from the negative charge toward the positive charge. On the external axial side of a dipole, the field contributions from the two charges combine along the dipole axis, and the resultant electric field points in the direction of the dipole moment. At a far point its magnitude is approximately proportional to 2p/(4πε₀r³), but the key fact here is its direction. Therefore option A is correct; it is not perpendicular or always zero.

An equatorial point lies on the perpendicular bisector of the dipole axis. At this point, the components of the fields due to the positive and negative charges perpendicular to the dipole axis cancel, while the components along the axis add in the direction opposite to the dipole moment. Thus the resultant field is antiparallel to p, with far-field magnitude approximately p/(4πε₀r³). Hence option A is correct.

The key directional ideas are different. In electric flux, the area vector is perpendicular to the surface; for a closed surface it is conventionally outward. In an electric dipole, the dipole moment points from the negative charge to the positive charge. Confusing these directions can produce an incorrect sign or angle in p·E or E·dA. Thus option A identifies the real mistake; the other options concern unrelated quantities.

Gauss’s law states that the net electric flux through a closed surface is Φ = Q_enclosed/ε₀. Here the enclosed charge is (+2 μC) + (−2 μC) = 0, so Φ = 0/ε₀ = 0. An external electric field can produce positive flux through some portions and negative flux through others, but these contributions cancel in the complete closed-surface integral.

For any closed surface, Gauss’s law gives Φ = Q_enclosed/ε₀, so only the algebraic charge inside the surface determines the net flux. Since the positive charge is enclosed and the negative partner lies outside, Q_enclosed is positive and the total flux is positive. The outside charge may alter the field distribution on the surface, but it does not contribute to net enclosed charge or replace the role of the enclosed charge.

A complete electric dipole consists of equal and opposite charges, +q and −q. If both charges lie inside the closed surface, the enclosed charge is Q_enclosed = q − q = 0. Gauss’s law therefore gives net flux Φ = Q_enclosed/ε₀ = 0, regardless of the surface shape. This does not mean the electric field vanishes at every point; only the total outward and inward flux contributions cancel.

Gauss’s law gives the total flux through a closed surface as Φ = Q_enclosed/ε₀. Therefore, if the net charge enclosed remains unchanged, the total flux remains unchanged even when the surface is stretched, compressed, or reshaped. The electric field and the flux density on individual portions may change, and the surface area need not remain constant, but the complete algebraic flux integral is fixed by enclosed charge.

For a uniform electric field through a plane surface, electric flux is Φ = EA cos θ, where θ is measured from the electric field to the area vector. At θ = 120°, cos 120° = −1/2, so Φ = −EA/2 for positive E and A. The negative sign indicates that the field component is opposite to the chosen area-vector direction; it is not zero because the angle is not 90°.

The electric flux through a plane surface in a uniform field is Φ = EA cos θ. Initially Φ₁ = EA cos 60° = 0.5EA. Finally Φ₂ = EA cos 30° = (√3/2)EA, which is about 0.866EA. Since √3/2 is greater than 0.5, the flux increases by a factor of √3, while its positive sign remains unchanged because both angles are less than 90°.

Electric flux through a plane surface is Φ = EA cos θ. When the electric field is at 90° to the area vector, cos 90° = 0, so the component of E normal to the surface is zero and Φ = 0, regardless of how large E is. The field may run parallel to the surface and still be strong, but it does not pass through the surface in the direction counted by the area vector.

For an 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. After the changes, q′ = 3q and d′ = d/3. Therefore p′ = q′d′ = (3q)(d/3) = qd = p. The two scaling factors cancel, so the dipole moment remains unchanged. Option A is correct.

The dipole moment is p = qd. Let the original values be q and d, so p = qd. If the new separation is d′ = d/2 and the new charge is q′ = kq, the required moment is p′ = 4p. Hence kq(d/2) = 4qd. Cancelling qd gives k/2 = 4, so k = 8. The charge magnitude must therefore become eight times its original value; option A is correct.

A polar molecule has separated centres of positive and negative charge, so it behaves as an electric dipole. In a non-uniform field, the forces on its two charges can have different magnitudes. Their unequal forces produce a net force, often toward the stronger-field region, while their different lines of action can produce torque that rotates the molecule. Therefore option A is correct; net positivity is not required, and options C and D contradict electrostatic principles.

Electric field lines are drawn in the direction of the force on a positive test charge; outside a dipole they therefore point from the positive charge toward the negative charge. The electric dipole moment is defined independently as p = qd, with its direction from the negative charge to the positive charge. These are different conventions describing different physical ideas, so there is no contradiction. Option A is correct; dipole moment is a vector, not a scalar.

At an equatorial point of a dipole, the electric field vector is antiparallel to the dipole moment: E_equatorial is opposite to p. Reversing the dipole moment changes p to −p. Since the field depends linearly on the dipole moment, the field also changes to −E, so its direction reverses while its magnitude at the same distance remains unchanged. Therefore option A is correct; it does not become zero or axial.

Electric flux through a plane surface is Φ = E·A = EA cos θ, where A is the area vector normal to the surface and θ is the angle between E and A. Therefore only the component of the electric field along the area vector, E cos θ, crosses the surface and contributes to flux. The component parallel to the surface gives zero dot product. Thus option A is correct; horizontal direction has no special role.

Flux through a plane surface is Φ = EA cos θ, where θ is the angle between the electric field E and the area vector A. The maximum value occurs when cos θ = 1, so the area vector must be parallel to the electric field. Because the area vector is normal, or perpendicular, to the surface itself, the surface must be perpendicular to the field. Hence option A is correct; a parallel surface gives zero flux.

An electric dipole consists of equal and opposite charges, +q and −q, so its net charge is zero. For any closed surface that completely encloses both charges, Gauss’s law gives Φ = Q_enclosed/ε₀ = (+q − q)/ε₀ = 0. Enlarging the surface does not change the enclosed net charge, although the local field and the distribution of entering and leaving lines may change. Hence option A is correct; flux is not equal to dipole moment.

Electric flux through a plane surface is given by Φ = EA cos θ, where θ is specifically the angle between the electric field and the area vector. Substituting E = 10 N/C, A = 4 m² and cos 60° = 1/2 gives Φ = 10 × 4 × 1/2 = 20 N m²/C. Thus option A is correct. Option B ignores the angular factor, option C omits the area, and option D would require a ninety-degree angle.

Flux is defined by Φ = E · A = EA cos θ, so its sign depends on the chosen orientation of the area vector. Reversing that vector changes A to −A and therefore changes Φ to −Φ. Since the electric field and surface area are unchanged, the magnitude remains the same while the sign becomes negative. Hence option A is correct; the other choices do not follow from the dot-product definition.

Gauss’s law states that the net electric flux through a closed surface is Φ = Q_enclosed/ε₀. Therefore only the algebraic sum of charges inside the surface determines the net flux. The external negative charge can alter the electric field at individual points on the surface, but its total contribution to closed-surface flux is zero because it is not enclosed. Thus option A is correct, while options B and C incorrectly include the outside charge.

For a closed surface, net flux is the algebraic sum of outward and inward contributions. Outward crossings count positively and inward crossings negatively. If equal amounts of field cross outward and inward, the total is zero even though the electric field is nonzero at many points. Therefore option A is correct. Zero net flux does not mean zero field everywhere, and electric flux itself is a scalar quantity obtained from a surface integral.

A cube has six identical faces, and a charge at its exact centre has no preferred direction relative to any face. By symmetry, Gauss’s law gives a total flux Q/ε₀, which is shared equally among the six faces. Hence flux through one face is (Q/ε₀)/6, or one sixth of the total. Option A is correct; one fourth would apply to four equivalent surfaces, while one half and the whole flux ignore the sixfold symmetry.

Gauss’s law fixes the total flux through a closed surface as Φ = Q_enclosed/ε₀. Moving the charge within the surface does not change the amount of enclosed charge, so the total flux remains unchanged. However, the field may become stronger over some regions and weaker over others, changing the local distribution of flux. Thus option A is correct; the other choices confuse local changes with the unchanged total.