Electric Flux and Electric Dipole

The flux through a plane surface is Φ = EA cosθ, where θ is the angle between the electric field and the chosen area vector. Maximum positive flux occurs when θ = 0°, so Φ = EA. Reversing the area vector changes θ to 180°, while E and A retain their magnitudes. Therefore Φ′ = EA cos180° = −EA, which is maximum negative flux. The magnitude remains maximum, but the sign reverses.

The governing relation for a plane surface in a uniform electric field is Φ = EA cos θ, where θ is the angle between the field and the area vector. Maximum flux is Φmax = EA when θ = 0°. Given Φ = Φmax/2, we get cos θ = 1/2, so θ = 60°. Therefore option A is correct. Ninety degrees gives zero flux, zero degrees gives maximum flux, and thirty degrees gives 0.866 times the maximum.

For a point far from an electric dipole, the field magnitude on the equatorial line varies as Eeq ∝ p/r³, where p is the dipole moment and r is the distance from its centre. If the distance changes from r to 2r, the new field is proportional to 1/(2r)³ = 1/(8r³). Thus E′ = E/8, so option A is correct. This is not an inverse-square dependence, which would have given one-fourth.

The dipole moment of two equal and opposite charges is defined by the vector relation p = qd, where q is the magnitude of either charge and d is the separation vector from the negative to the positive charge. Hence p becomes zero if q = 0 or d = 0. A uniform or non-uniform external field does not define whether the intrinsic moment is zero. The field ratio in option D is a property of a far dipole field, not a condition for p = 0. Therefore A is correct.

Electric flux through a plane surface is Φ = EA cos θ, where θ is measured between the electric field and the area vector, which is perpendicular to the surface. If the field is parallel to the surface, then it is perpendicular to the area vector, so θ = 90° and cos 90° = 0. Doubling E still gives Φ = 2EA(0) = 0. Hence option A is correct; changing magnitude cannot create a normal component when direction is unchanged.

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. The north-east field has a component in the northward direction, so θ is less than 90° and cos θ is positive. Hence the flux is positive. It is not zero and is not maximum because the field is not exactly parallel to the area vector.

Electric flux is defined by the scalar product Φ = E·A, so its result is a scalar quantity, although its sign depends on the chosen surface orientation. Electric dipole moment is p = qd; it has a magnitude and a definite direction from the negative charge toward the positive charge, so it is a vector. Therefore option A correctly states the distinction. Options B and C classify one quantity incorrectly, while D gives incorrect definitions.

Electric flux is given by Φ = E A cos θ, where θ is measured from the area vector to the electric field. For θ = 120°, cos 120° = −1/2. Since E and A are positive magnitudes, Φ = −EA/2, which is negative. Therefore option A is correct. Positive flux occurs for an angle below 90°, zero flux at 90°, and maximum positive flux at 0°, so the other choices do not fit this angle.

For a uniform field and unchanged area, electric flux is Φ = E A cos θ. Initially, Φ₁ = EA cos 30° = (√3/2)EA. Finally, Φ₂ = EA cos 60° = (1/2)EA. Since 1/2 is less than √3/2, the flux decreases while remaining positive. Option A is correct. It does not become negative because both angles are below 90°, and the field and area have not changed.

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. If q becomes 4q and d becomes d/2, then the new moment is p′ = (4q)(d/2) = 2qd = 2p. Therefore the dipole moment becomes double, so A is correct. B considers only the changed separation, C considers only the changed charge, and D ignores both changes.

The dipole moment is governed by p = qd. Let the initial values be p = qd. After the changes, let q become kq and d become 3d. The required moment is 9p, so (kq)(3d) = 9(qd). Cancelling qd gives 3k = 9 and hence k = 3. The charge magnitude must therefore be tripled. Option A is correct; B overlooks the threefold increase in separation, while C and D give the wrong scaling.

Electric flux through a surface is defined by Φ = ∫ E · dA. Reversing the area vector changes dA to −dA while the electric field remains unchanged, so the entire integral changes from Φ to −Φ. Thus a negative flux becomes positive with the same magnitude, not zero and not double. This sign reversal is associated with changing the chosen orientation of an open surface; it does not imply that the physical field has changed. Hence A is correct.

A polar molecule has separated positive and negative charge centres and behaves as an electric dipole. In a uniform field, the forces on its two charges can be equal and opposite, producing no net force. In a non-uniform field, the field magnitudes at the two ends are different, so F+ = qE+ and F− = qE− do not cancel completely. A net force can therefore act, in addition to possible torque.

In a non-polar molecule, the centres of positive and negative charge normally coincide, so its permanent dipole moment is zero. An applied electric field exerts opposite effects on the nucleus and electron cloud, causing a small displacement between these charge centres. This induced separation produces an induced dipole moment, usually proportional to the applied field, p = αE. The molecule remains electrically neutral; its net charge does not become positive.

Net charge and dipole moment are different physical quantities. Net charge is the algebraic sum of all charges, so +q and −q give zero net charge. The dipole moment of two equal and opposite charges is p = qd, where d is the separation vector from negative to positive charge. Hence a dipole can have zero net charge but a non-zero moment whenever q and d are both non-zero. Separation is essential, not irrelevant.

Gauss’s law states that the total flux through a closed surface is Φ = Qenclosed/ε0. A complete dipole has charges +q and −q, so its enclosed net charge is zero and the total flux is zero. However, flux is the surface integral of the normal component of electric field, not the field value at each point. Contributions entering and leaving different parts can cancel, while the local electric field remains non-zero.

The electric dipole moment is defined as p = qd, with its direction from the negative charge to the positive charge. Electric field lines, however, indicate the direction of force on a positive test charge; outside an electric dipole they run from the positive charge toward the negative charge. Thus the dipole-moment direction and the external field-line direction are opposite along the dipole axis. Confusing these conventions leads to incorrect diagrams and force or torque conclusions.

The governing relation for an electric dipole is p = qd, where q is the magnitude of either charge and d is the separation vector from the negative to the positive charge. If the opposite charges are non-zero and separated by a non-zero distance, both q and d are non-zero, so p cannot be zero. Therefore option A is correct; the other options incorrectly ignore charge magnitude or separation.

Electric flux is defined by the surface integral Φ = ∫ E·dA, a scalar (dot) product. The result is therefore a scalar, but it may carry a sign: for a uniform field, Φ = EA cosθ. It is positive when E has a component along the chosen area vector and negative when it points oppositely. Thus option A is correct; the sign does not make flux a vector and flux does have units, N·m²/C.

For an open surface, the area vector dA is assigned by convention and has a chosen normal direction. Flux is Φ = ∫ E·dA, so reversing the normal changes dA to −dA and consequently changes Φ to −Φ, while the physical field and surface area remain unchanged. Therefore option A is correct. The sign change is a convention linked to orientation, not evidence that charge or the electric field has disappeared or changed.

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. Substituting E = 8 N/C, A = 6 m², and cos 60° = 1/2 gives Φ = 8 × 6 × 1/2 = 24 N·m²/C. Thus option A is correct. Option B ignores the angular factor, while zero would require a 90° angle.

Electric flux is the dot product Φ = E · A. Reversing the area vector changes A to −A, so the new flux becomes Φ′ = E · (−A) = −Φ. The magnitude is unchanged because the field and surface area have not changed; only the chosen orientation has changed. Therefore, a positive initial flux becomes a negative flux of equal magnitude, making option A correct.

Gauss’s law states that net electric flux through a closed surface is Φ = Q_enclosed/ε₀. Initially, only the positive charge is enclosed, so Q_enclosed is positive and the flux is positive. After enlargement, equal positive and negative charges are both enclosed; their algebraic sum is zero. Therefore the net flux changes from positive to zero. The field may still exist locally, but the total flux is zero.

For a surface element, flux is defined by dΦ = E · dA = E dA cos θ. A closed surface has a natural outward normal, but an open surface has no unique inside or outside, so its normal direction must be selected by convention. Reversing the chosen area vector changes the sign of the calculated flux, although the physical surface and field remain unchanged. Therefore option A is correct.

Flux measures the component of electric field passing through a surface. From Φ = EA cos θ, the factor E cos θ is the component normal to the surface. The component parallel to the surface has θ = 90° relative to the normal, so its contribution is E_parallel A cos 90° = 0; it merely runs along the surface. Thus only the normal component crosses the surface and contributes to flux.