Electric Flux and Electric Dipole

For a uniform field through a plane surface, Φ = EA cos θ. Since E and A remain constant, compare only the cosine factors: cos 30° = √3/2, whereas cos 60° = 1/2. Therefore Φ₂/Φ₁ = (1/2)/(√3/2) = 1/√3, which is smaller than one. The flux decreases, even though it does not become zero at sixty degrees.

Electric field lines in the space between the charges point from the positive charge toward the negative charge. The electric dipole moment is defined independently as p = qd, with the vector d directed from the negative charge to the positive charge. Therefore the dipole moment points from negative to positive, opposite to the field direction in the central region. It is a vector and does have a definite direction.

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. If q becomes 2q and d becomes 2d, then p_new = (2q)(2d) = 4qd = 4p. Thus the dipole moment becomes four times its original magnitude. The opposite signs of the charges remain essential, but they do not change this scaling result.

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. Initially p = qd. After the changes, q' = q/2 and d' = 4d, so p' = (q/2)(4d) = 2qd = 2p. Therefore, the dipole moment becomes double. It is not merely half or four times because both changes must be combined.

A permanent electric dipole moment requires a lasting separation between the centres of positive and negative charge. In a non-polar molecule, the charge distribution is symmetric in the absence of an external field, so the two charge centres effectively coincide. Using p = qd, the effective separation d is zero and hence p = 0. Such a molecule may acquire an induced dipole moment in an external field, but that is not a permanent moment.

Electric flux through a plane surface 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 not perpendicular to the surface, only the normal component E cos θ contributes to flux. A common mistake is to use Φ = EA without the cosine factor, effectively treating the entire field as perpendicular. Including area and its direction is necessary, not an error.

Electric flux through a plane surface is given by Φ = EA cos θ, or more generally by the product of electric field, area, and the angular factor. The changed factors are 2E, 2A, and (cos θ)/2. Therefore, Φ′/Φ = 2 × 3 × 1/2 = 3. Hence the new flux is three times the original value. Six times would result from ignoring the halving factor.

Electric flux is Φ = EA cos θ, where θ is the angle between the electric field and the area vector. When the surface is perpendicular to E, its area vector is parallel to E, so θ = 0° and cos θ = 1. This is the largest possible value of the angular factor, giving Φmax = EA. If the field ran along the surface, θ would be 90° and the flux would be zero.

The governing concept is the distinction between a field at a point and the flux through a surface. Electric field E has magnitude and direction, so it is a vector. Electric flux is the surface integral Φ = ∫E·dA; its value measures the net field passing through a chosen surface and is scalar. Therefore A is correct. B wrongly calls flux always a vector, while C and D do not describe either physical quantity.

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. Here θ = 120°, and cos 120° = −1/2. Hence Φ = −EA/2, so the flux is negative. It is not zero, because the angle is not 90°, and it is not positive because the cosine is negative.

For a plane surface, Φ = EA cos θ. Maximum positive flux occurs when the area vector is parallel to the electric field, so θ = 0° and Φ = EA. Reversing the surface reverses its area vector, making θ = 180°. Therefore Φ = EA cos 180° = −EA: the magnitude remains maximum, but the sign becomes negative. Hence A is correct.

The flux relation is Φ = EA cos θ, while the maximum positive flux is Φ_max = EA. If Φ = Φ_max/2, then EA cos θ = EA/2, giving cos θ = 1/2. For the usual angle between vectors, θ = 60°. At 0° the flux is maximum, at 90° it is zero, and at 180° it is maximum negative. Therefore A is correct.

For a plane surface, flux is Φ = EA cos θ, where θ is the angle between the field and the area vector. If the electric field is parallel to the surface, it is perpendicular to the area vector, so θ = 90° and cos 90° = 0. Thus the flux is zero even though E is non-zero. A perpendicular field would give maximum magnitude, so B and C are incorrect.

The magnitude of the 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 the charge becomes 5q, keeping p unchanged requires (5q)d′ = qd. Dividing by 5q gives d′ = d/5. Therefore option A is correct. Making separation five times would increase p, while unchanged or half separation would not compensate for the fivefold charge increase.

For an electric dipole, the moment magnitude is p = qd. Let the original values be q and d, so p = qd. The changed values are q′ = 4q and d′ = d/2. Hence p′ = q′d′ = (4q)(d/2) = 2qd = 2p. The dipole moment therefore becomes double, making option A correct. It is not four times because the halved separation reduces the effect by a factor of two.

For a point far from a short electric dipole, the axial field varies as E_axial ∝ 1/r³. If the distance changes from r to 2r, the new field is E′ = E(r/2r)³ = 72/8 = 9. Thus option A is correct. The value 36 would correspond to an inverse-square dependence, while 18 represents an incorrect factor of four reduction; 144 would imply an increase rather than the required decrease.

In the far-field region of an electric dipole, the magnitude of the equatorial field also follows E_equatorial ∝ 1/r³. When the distance changes from r to 3r, the field is reduced by 3³ = 27. Therefore E′ = 27/27 = 1 unit, so option A is correct. Option C would result from using an inverse-square law, and option B does not apply the cubic dependence. Option D has the wrong direction of change.

For a short dipole observed at the same far distance, the axial field magnitude is E_axial = 2kp/r³, whereas the equatorial field magnitude is E_equatorial = kp/r³. Hence the axial field is twice the equatorial field. Given E_axial = 54, we obtain E_equatorial = 54/2 = 27. Therefore option A is correct. The axial value itself ignores the factor of two, while 108 reverses the relation.

At equal large distances from a short electric dipole, the magnitudes are E_axial = 2kp/r³ and E_equatorial = kp/r³. Thus the axial field is exactly twice the equatorial field. With E_equatorial = 16, E_axial = 2 × 16 = 32. Therefore option A is correct. Eight is half the given value, sixteen incorrectly assumes equality, and sixty-four applies an unjustified factor of four.

Electric flux through a plane surface is Φ = EA cos θ, where θ is measured between the electric field and the area vector. At θ = 90°, cos θ = 0. If θ is slightly less than 90°, cos θ is small but positive, so the flux is positive and has a small magnitude, assuming E and A are nonzero. It is not maximum, because maximum positive flux occurs at θ = 0°. Therefore A is correct.

The flux through a surface is Φ = EA cos θ. For an angle just greater than 90° but less than 180°, the cosine is negative. With nonzero E and area A, the product EA cos θ is therefore negative. Its magnitude may be small when the angle is only slightly above 90°, but the question asks for the sign. Hence A is correct; the flux is zero only at exactly 90°, and it is positive for angles below 90°.

Electric flux is the scalar product Φ = E·A, or more generally ∫E·dA. Its value is scalar, but the dot product can be positive, zero, or negative according to the angle between the electric field and the chosen area vector. Negative flux means the field component is opposite to that oriented area vector; it does not make flux a vector and does not mean that area is negative.

The electric dipole moment is defined as p = qd, directed from the negative charge to the positive charge, so it is a vector with an intrinsic direction. Electric flux is the scalar product E·A; its positive or negative sign records the orientation of the field relative to the chosen area vector. Therefore flux has no independent vector direction, making option A the correct comparison.

Electric flux through a plane surface is Φ = EA cos θ. With θ unchanged, cos θ is constant, so flux is proportional to the product EA. The area changes from A to 3A and the field changes from E to E/2. Therefore Φ' = (3A)(E/2)cos θ = 3Φ/2. The flux becomes one and a half times, so option A is correct.

Electric flux is Φ = EA cos θ, where θ is measured between the electric field and the area vector. Since E and A remain fixed, compare the cosine factors: cos 30° = √3/2, whereas cos 60° = 1/2. The first value is larger, so the flux increases when the angle changes from 60° to 30°. Therefore option A is correct; it does not become zero or negative.