Electric Flux and Electric Dipole

Flux through a surface is Φ = EA cos θ. A negative value means the field’s directional component relative to the chosen area vector gives a negative sign. If the field magnitude is doubled while its direction, the area, and orientation remain unchanged, the entire flux is multiplied by two: Φ' = 2Φ. Because the original Φ is negative, the new flux is still negative but has twice its magnitude. Thus option A is correct.

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 the changes, p' = 2p and d' = 2d. Substituting gives 2p = q'(2d), so q'd = qd and q' = q. Therefore the charge magnitude is unchanged, making option A correct.

For a two-charge electric dipole, the dipole moment magnitude is p = qd. Let the original values be p = qd. If the new moment is 3p and the new separation is 3d, then 3p = q'(3d). Cancelling the factor of three gives p = q'd, and comparison with p = qd yields q' = q. Hence the charge magnitude remains unchanged, so option A is correct.

The governing concept is that electric-field lines show the direction in which a positive test charge would move, so outside an electric dipole they are conventionally drawn from the positive charge toward the negative charge. Electric dipole moment is defined separately as p = q(2a), directed from the negative charge to the positive charge. Therefore option A correctly distinguishes the two directions; options B and C incorrectly make them identical, while D ignores their vector nature.

The governing idea is that electric flux is calculated as Φ = E·A, where the area vector A is perpendicular to the chosen surface and its orientation determines the sign of flux. Dipole moment, however, is defined from the negative charge to the positive charge. These directions need not coincide. Thus option A is the essential caution; option C is wrong because flux can be negative, and B and D have no valid physical basis.