Electric Flux and Electric Dipole

An electric dipole consists of charges +q and −q. If the closed surface encloses the complete dipole, its net enclosed charge is Q_enclosed = +q − q = 0. Gauss’s law therefore gives Φ = Q_enclosed/ε₀ = 0. The electric field is not necessarily zero at every point on the surface; only the total flux is zero. Hence the correct choice is A.

For any closed surface, Gauss’s law gives Φ = Q_enclosed/ε₀, so only the charge actually inside is counted. Here the positive charge +q is enclosed, while the negative partner lies outside and contributes zero net flux to the closed surface, even though it creates a field on the surface. Therefore Φ = +q/ε₀ and depends on the enclosed positive charge.

For a uniform electric field through a plane surface, electric flux is Φ = EA cos θ, where E is field magnitude, A is area, and θ is the angle between the field and area vector. With E and θ fixed, Φ is directly proportional to A. Replacing A by 2A gives Φ′ = E(2A)cosθ = 2Φ. Therefore the flux doubles.

The flux through a plane surface is Φ = EA cos θ. When area A and angle θ remain unchanged, flux is directly proportional to electric-field magnitude E. If E becomes 2E, then Φ′ = (2E)A cosθ = 2Φ. Thus the flux doubles. It does not become half or zero, and “four times smaller” has no basis because no inverse-square dependence applies here.

Electric flux is defined by the dot product Φ = E·A = EA cos θ. The area vector specifies the chosen normal direction. Flux is positive when the electric field has a component along that direction, meaning θ is less than 90° for a simple uniform case. Positive flux does not mean the field is absent or that area is zero; inward field components would instead give negative flux.

Flux is the dot product Φ = E·A = EA cos θ, with the area vector defining the chosen positive normal. If the electric field points opposite to that vector, θ is greater than 90°, so cos θ is negative and the flux is negative. The sign indicates direction relative to the chosen normal, not merely a small magnitude. A field along the vector gives positive flux.

An electric dipole has equal and opposite charges +q and −q separated by distance d. Its dipole-moment magnitude is p = qd, where q is the magnitude of either charge and d is the separation. The vector points from the negative charge to the positive charge. Although the net charge q + (−q) is zero, the dipole moment is generally nonzero; therefore option A is correct.

The electric dipole moment is defined as p = qd, where q is the magnitude of either charge and d is the separation vector directed from the negative charge to the positive charge. If q becomes zero, multiplication gives p = 0 × d = 0, regardless of the separation. Therefore, option A is correct. A nonzero separation alone cannot produce a dipole moment without separated charges.

For an electric dipole, the dipole moment is expressed as p = qd, where q is the charge magnitude and d is the separation between the opposite charges. When the separation becomes zero, p = q × 0 = 0. Thus the two charges no longer form a finite separated dipole. Option A is correct; doubling and infinity do not follow from d = 0, and the moment is a vector rather than always negative.

An electric dipole consists of two equal and opposite charges separated by a small distance. That distance is commonly called the separation of charges, usually represented by d in the expression p = qd for dipole moment. Hence option A is correct. Net charge describes the algebraic sum of charges, current means charge flow per unit time, and an area vector represents a surface’s area and orientation.

Electric flux through a plane surface is Φ = EA cos θ for a uniform field, where E is the field magnitude, A is the area, and θ is the angle between the electric field and the area vector normal to the surface. Rotation changes this angle, so cos θ and therefore flux can change even when E and A remain constant. Thus option A is correct; mass, units, and charge disappearance are irrelevant.

For a uniform field, flux is Φ = EA cos θ, where θ is measured between the field and the area vector, which 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; consequently Φ = 0. Therefore option A is correct. A perpendicular field generally gives maximum flux, while large area or strong field increases magnitude.

By Gauss’s law, the net outward electric flux through a closed surface is proportional to the net charge enclosed: Φ = Q_enclosed/ε₀. More field lines leaving than entering indicate positive net outward flux. Therefore Q_enclosed is positive, so option A is correct. A negative charge would produce more inward lines, while equal inward and outward contributions would indicate zero net charge, assuming the diagram represents field-line density consistently.

For a closed surface, outward flux is taken as positive and inward flux as negative. If the represented number and strength of lines entering and leaving are equal, the positive and negative contributions cancel. Hence the net electric flux is zero, which makes option A correct. This also corresponds, through Gauss’s law, to zero net enclosed charge, although individual charges may still be present if their algebraic sum is zero.

For a plane surface in a uniform electric field, Φ = EA cos θ, where θ is the angle between the field and the area vector. The flux is maximum when cos θ = 1, meaning the area vector is parallel to the field. Since the area vector is perpendicular to the surface, the surface itself must be perpendicular to the electric field. Therefore option A is correct; a parallel surface gives zero flux.

The electric dipole moment of two equal and opposite charges is given by p = qd, where q is the magnitude of either charge and d is the separation between them. Therefore, increasing q or d increases p, provided the other quantity remains non-zero. Options B and D make the separation zero, while option C makes the charge zero; in each case the dipole moment becomes zero. Hence option A is correct.

Direction is important in both ideas, although it appears through different relations. Electric flux is Φ = EA cos θ, so the angle between the electric field and the area vector changes the flux. A dipole has a directed dipole-moment vector, and its orientation relative to an external field affects torque and potential energy. Mass and colour are irrelevant, and neither quantity is always zero. Thus option A is correct.

For a uniform electric field, flux through a plane surface is Φ = EA cos θ, where E is field strength, A is area, and θ is the angle between the field and the area vector. Since E and θ are unchanged, doubling A multiplies Φ by two. Therefore option A is correct. It would become zero only at 90° to the area vector, not merely because area changes.

Electric flux through a plane surface is Φ = EA cos θ. Because the orientation is unchanged, cos θ remains constant. The new flux is Φ′ = (2E)(A/2)cos θ = EA cos θ = Φ. The doubling of field strength and halving of area exactly cancel, so option A is correct. Options B and C consider only one change, while D would require a perpendicular field to the area vector.

Electric flux is Φ = EA cos θ. Maximum flux for fixed E and A occurs when θ = 0°, so Φmax = EA. At θ = 60°, Φ = EA cos 60° = EA × 1/2 = Φmax/2. Hence option A is correct. Zero flux occurs at 90°, while full maximum flux occurs at 0°; doubling is impossible without changing E or A.

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 changes, p′ = (2q)(d/2) = qd = p. Thus the dipole moment remains unchanged in magnitude, although its direction would still be from the negative charge toward the positive charge. Option A is correct.

For an electric dipole, the dipole moment magnitude is p = qd. Since the charge q is held constant, p is directly proportional to the separation d. If the required moment is p′ = 3p, then qd′ = 3qd. Dividing by the unchanged q gives d′ = 3d. Therefore the separation must be tripled, making option A correct; reducing or leaving d unchanged cannot produce the required increase.

The dipole moment magnitude is p = qd. Let the original values be q and d, so p = qd. The new charge is q′ = q/2 and the new separation is d′ = 4d. Hence p′ = q′d′ = (q/2)(4d) = 2qd = 2p. Therefore the dipole moment becomes double. Option A is correct; option B considers only the charge change, while C considers only the separation change.

Electric flux through a plane surface is Φ = EA cos θ, where θ is the angle between the electric field and the area vector. Eastward and westward directions are opposite, so θ = 180° and cos 180° = −1. Hence Φ = −EA, which is negative for nonzero field and area. Option A is correct. Flux would be positive for the same direction, zero for perpendicular vectors, and maximum positive when θ = 0°.

The relevant relation is Φ = EA cos θ. Northward and eastward directions are perpendicular, so the angle between the electric field and the area vector is 90°. Since cos 90° = 0, the scalar electric flux is zero, regardless of the nonzero values of E and A. Therefore option A is correct. Maximum positive flux requires parallel vectors, maximum negative flux requires opposite vectors, and half maximum would require a 60° angle in magnitude.